4 Linear Dierential Equations


 Arleen Moore
 3 years ago
 Views:
Transcription
1 Dierential Equations (part 2): Linear Dierential Equations (by Evan Dummit, 2012, v. 1.00) Contents 4 Linear Dierential Equations Terminology General Theory of Linear Dierential Equations Homogeneous Linear Equations with Constant Coecients NonHomogeneous Linear Equations with Constant Coecients Undetermined Coecients Variation of Parameters SecondOrder Equations: Applications to Newtonian Mechanics Spring Problems and Damping Resonance and Forcing Linear Dierential Equations The general nthorder linear dierential equation is of the form y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = Q(x), for some functions P n (x),, P 2 (x), P 1 (x), and Q(x). (Note that y (n) denotes the nth derivative of y.) The goal is to study the behavior of the solutions to these equations. Solving general linear dierential equations explicitly is generally hard, unless we are lucky and the equation has a particularly nice form. We can only give a method for writing down the full set of solutions for a small classes of linear equations: namely, linear dierential equations with constant coecients. There are a few equation types (e.g., Euler equations like x 2 y + xy + y = 0) which can be reduced to constantcoecient equations via substitution. Thus, we will spend most of our eort discussing how to solve linear equations with constant coecients, because we can always solve them these are equations of the form y (n) +a n 1 y (n 1) + +a 1 y +a 0 y = Q(x) for some constants a n 1,, a 0 and some function Q(x). Even harder than the general linear dierential equation are nonlinear equations of higher order, such as y = e y y or y y 2 = y + 1. We will not discuss nonlinear equations at all. Compare with the example of rstorder equations: we can solve any rstorder linear equation, but very few types of rstorder nonlinear equations. 4.1 Terminology The standard form of a dierential equation is when it is written with all terms involving y or higher derivatives on one side, and functions of the variable on the other side. Example: The equation y + y + y = 0 is in standard form. Example: The equation y = 3x 2 xy is not in standard form.
2 An equation is homogeneous if, when it is put into standard form, the xside is zero. nonhomogeneous otherwise. An equation is Example: The equation y + y + y = 0 is homogeneous. Example: The equation y + xy = 3x 2 is nonhomogeneous. An nth order dierential equation is an equation in which the highest derivative is the nth derivative. Example: The equations y + xy = 3x 2 and y y = 2 are rstorder. Example: The equation y + y + y = 0 is secondorder. A dierential equation is linear if it is a linear combination of y and its derivatives. (Note that coecients are allowed to be functions of x.) In other words, if there are no terms like y 2, or (y ) 3, or y y, or e y. Example: The equations y + xy = 3x 2 and y + y + y = 0 are linear. Example: The equations y y = 3x 2 and y + e y = 0 are not linear. We say a linear dierential equation has constant coecients if the coecients of y, y, y,... are all constants. Example: The equation y + y + y = 0 has constant coecients. Example: The equation y + xy = 3x 2 does not have constant coecients. For n functions y 1, y 2,, y n which are each dierentiable (n 1) times, the Wronskian W (y 1, y 2,, y n ) of y 1 y 2 y n y 1 y 2 y n the functions is dened to be the determinant of the matrix y (n 1) 1 y (n 1) 2 y n (n 1) Note that the Wronskian will also be a function of x. The purpose of the Wronskian is to provide a way to show that functions are linearly independent. Theorem: A collection of n, ntimes dierentiable functions y 1, y 2,, y n is linearly independent (in the vector space of ntimes dierentiable functions) if their Wronskian is not the zero function. To prove the theorem, note that if the functions are linearly dependent with dierentiating the appropriate number of times we see that n j=1 a j y (i) j n a j y j = 0, then by j=1 = 0 for any 0 i n. Hence, in particular, the rows of the matrix are linearly dependent (as vectors), and so the determinant of the matrix is zero. Therefore, if the determinant is not zero, the functions cannot be dependent. Remark: The theorem becomes an ifandonly if statement (i.e., the functions are linearly independent if and only if the Wronskian is nonzero) if we know that the functions y 1, y 2,, y n are innitely dierentiable. The proof of the other direction is signicantly more dicult. Example: The functions 1 and x are linearly independent, because we can compute W (1, x) = 1 x 0 1 = 1. Example: The functions sin(x) and cos(x) are linearly independent, as W (sin(x), cos(x)) = sin(x) cos(x) cos(x) sin(x) 1. Example: The functions 1, x, and 1+x are (rather clearly) linearly dependent. We have W (1, x, 1+x) = 1 x x = 0, by expanding along the bottom row =
3 4.2 General Theory of Linear Dierential Equations Theorem (Homogeneous Linear Equations): If P n (x),, P 1 (x) are continuous functions, then the set of solutions to the homogeneous nth order equation y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = 0 is an ndimensional vector space. The fact that the set of solutions is a vector space is not so hard to show using the subspace criteria. The real result of this theorem, which is analogous to the existenceuniqueness theorem for rstorder equations, is that the set of solutions is ndimensional. Example: Consider the homogeneous equation y (x) = 0. We can just integrate twice to see that the solutions are y(x) = Ax + B, for any constants A and B. Indeed, as the theorem dictates, the solution functions form a twodimensional space, spanned by the two basis elements 1 and x. Theorem (ExistenceUniqueness for Linear Equations): If P n (x),, P 1 (x) and Q(x) are functions continuous on an interval containing a, then there is a unique solution (possibly on a smaller interval) to the initial value problem y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = Q(x), for any initial conditions y(a) = b 1, y (a) = b 2,, and y (n 1) (a) = b n. Additionally, every solution y gen to the general nth order equation may be written as y gen = y par + y hom, where y par is any one particular solution to the equation, and y hom is a solution to the homogeneous equation y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = 0. What this theorem says is: in order to solve the general equation y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = Q(x), it is enough to nd one solution to this equation along with the general solution to the homogeneous equation. The existenceuniqueness part of the theorem is hard, but the second part is fairly simple to show: y 1 and y 2 are solutions to the general equation, then their dierence y 1 y 2 is a solution to the homogeneous equation to see this, just subtract the resulting equations and apply derivatives rules. A more advanced way to see the same thing is to use the fact that the map L sending y to y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y is a linear transformation. Then L(y 1 ) = L(y 2 ) says L(y 1 y 2 ) = 0, so that y 1 y 2 is a solution to the homogeneous equation. Example: In order to solve the equation y (x) = e x, the theorem says we only need to nd one function which is a solution, and then solve the homogeneous equation y (x) = 0. We can just try simple functions until we discover that y(x) = e x has y (x) = e x. Then we need only solve the homogeneous equation y (x) = 0, whose solutions we know are Ax + B. Thus the general solution to the general equation y (x) = e x is y(x) = e x + Ax + B. We can also verify that if we impose the initial conditions y(0) = c 1 and y (0) = c 2, then (as the theorem dictates) there is the unique solution y = e x + (c 2 1)x + (c 1 1). 4.3 Homogeneous Linear Equations with Constant Coecients The general linear homogeneous dierential equation with constant coecients is y (n) + a n 1 y (n 1) + + a 1 y + a 0 y = 0, where a n 1,, a 0 are some constants. From the existenceuniqueness theorem we know that the set of solutions is an ndimensional vector space. Based on solving rstorder linear homogeneous equations (i.e., y + ky = 0), we might expect the solutions to involve exponentials. If we try setting y = e rx then after some arithmetic we end up with r n e rx + a n 1 r n 1 e rx + + a 1 re rx + a 0 e rx = 0. Multiplying both sides by e rx and cancelling yields the characteristic equation r n + a n 1 r n a 1 r + a 0 = 0.
4 If we can nd n values of r satisfying this nthdegree polynomial i.e., if we can factor the polynomial and see that it has n distinct roots, then the theorem tells us we will have found all of the solutions. If we are unlucky and the polynomial has a repeated root, then we need to try something else. If there are nonreal roots (note they will come in complex conjugate pairs!) r 1 = α + βi and r 2 = α βi then we would end up with e r1x and e r2x as our solutions. But we really want realvalued solutions, and e r1x and e r2x have complex numbers in the exponents. However we can just write out the real and imaginary parts using Euler's Theorem, and take linear combinations to obtain the two realvalued solutions e αx sin(βx) = 1 2i [er1x e r2x ] and e αx cos(βx) = 1 2 [er1x + e r2x ]. Taking motivation from the case of y (k) = 0, whose characteristic equation is r k = 0 (with the kfold repeated root 0) and whose solutions are y(x) = A 1 + A 2 x + A 3 x A k x k 1, we guess wildly that if other roots are repeated, we want to multiply the corresponding exponentials e rx by a power of x. If we put all of these ideas together we can prove that this general outline will, in fact, give us n linearly independent functions, and hence gives the general solution to any homogeneous linear dierential equation with constant coecients. Advanced Remark: Another way to organize this information is to think of everything in terms of linear transformations. If D represents the linear transformation sending a function to its derivative, then we are looking for functions y with the property that L(y) = 0, where L is the linear operator D n + a n 1 D n a 1 D + a 0. Note that D 2, for instance, means taking the derivative twice in a row i.e., D 2 is the second derivative. Then what we are secretly doing is factoring the polynomial D n +a n 1 D n 1 + +a 1 D +a 0 into a product of linear terms (D r 1 ) k1 (D r j ) kj, solving each of the simpler equations (D r i ) ki y = 0, and adding up the resulting terms. To solve a linear homogeneous dierential equation with constant coecients, follow these steps: Step 1: Rewrite the dierential equation in the standard form y (n) + a n 1 y (n 1) + + a 1 y + a 0 y = 0 (if necessary). Step 2: Factor the characteristic equation r n + a n 1 r n a 1 r + a 0 = 0. Step 3: For each irreducible factor in the characteristic equation, write down the corresponding terms in the solution: For terms (r α) k where is a real number then add the terms of the form e αx, xe αx,, x k 1 e αx. For irreducible terms (r 2 + cx + d) k with roots r = α ± βi, then add the terms of the form e αx sin(βx), xe αx sin(βx),, x k 1 e αx sin(βx) and e αx cos(βx), xe αx cos(βx),, x k 1 e αx cos(βx). Step 4: If given additional conditions on the solution, solve for the coecients (if necessary). Example: Find all functions y such that y + y 6 = 0. Step 2: The characteristic equation is r 2 + r 6 = 0 which has roots r = 2 and r = 3. Step 3: We have two distinct real roots, so our terms are e 2x and e 3x. y = C 1 e 2x + C 2 e 3x. Example: Find all functions y such that y 2y + 1 = 0, with y(0) = 1 and y (0) = 2. Step 2: The characteristic equation is r 2 2r + 1 = 0 which has only the solution r = 1. So the general solution is Step 3: There is a double root at r = 1, so our terms are e x and xe x. Hence the general solution is y = C 1 e x + C 2 xe x. Step 4: Plugging in the two conditions gives 1 = C 1 e 0 + C 2 0, and 2 = C 1 e 0 + C 2 [ (0 + 1)e 0 ] from which C 1 = 1 and C 2 = 1. Hence the particular solution requested is y = e x + xe x. Example: Find all realvalued functions y such that y = 4y. Step 1: The standard form here is y + 4y = 0.
5 Step 2: The characteristic equation is r = 0 which has roots r = 2i and r = 2i. Step 3: We have two complexconjugate roots. Since the problem asks for realvalued functions we write e r1x = cos(2x) + i sin(2x) and e r2x = cos(2x) i sin(2x) to see that the general solution is y = C 1 cos(2x) + C 2 sin(2x). Example: Find all realvalued functions y such that y (5) + 5y (4) + 10y + 10y + 5y + y = 0. Step 2: The characteristic equation is r 5 + 5r r r 2 + 5r + 1 = 0 which factors as (r + 1) 5 = 0. Step 3: We have a 5fold repeated root r = 1. Thus the terms are e x, xe x, x 2 e x, x 3 e x, and x 4 e x. Hence the general solution is y = C 1 e x + C 2 xe x + C 3 x 2 e x + C 4 x 3 e x + C 5 x 4 e x. Example: Find all realvalued functions y whose fourth derivative is the same as y. Step 1: This is the equation y = y, or in standard form, y y = 0. Step 2: The characteristic equation is r 4 1 = 0 which factors as (r + 1)(r 1)(r + i)(r i) = 0. Step 3: We have the four roots 1, 1, i, i. Thus the terms are e x, e x, sin(x), and cos(x). Hence the general solution is y = C 1 e x + C 2 e x + C 3 sin(x) + C 4 cos(x). 4.4 NonHomogeneous Linear Equations with Constant Coecients The general linear dierential equation with constant coecients is of the form y (n) +a n 1 y (n 1) + +a 1 y + a 0 y = Q(x), where a n 1,, a 0 are some constants and Q(x) is some function of x. From the general theory, all we need to do is nd one solution to the general equation, and nd all solutions to the homogeneous equation. Since we know how to solve the homogeneous equation in full generality, we just need to develop some techniques for nding one solution to the general equation. There are essentially two ways of doing this. The Method of Undetermined Coecients is just a fancy way of making an an educated guess about what the form of the solution will be and then checking if it works. It will work whenever the function Q(x) is linear combination of terms of the form x k e αx (where k is an integer and α is a complex number): thus, for example, we could use the method for something like Q(x) = x 3 e 8x cos(x) 4 sin(x) + x 10 but not something like Q(x) = tan(x). Variation of Parameters is a more complicated method which uses some linear algebra and cleverness to use the solutions of the homogeneous equation to nd a solution to the nonhomogeneous equation. It will always work, for any function Q(x), but generally requires more setup and computation Undetermined Coecients The idea behind the method of undetermined coecients is that we can 'guess' what our solution should look like (up to some coecients we have to solve for), if Q(x) involves sums and products of polynomials, exponentials, and trigonometric functions. Specically, we try a solution y = [stu], where the 'stu' is a sum of things similar to the terms in Q(x). Here is the procedure for generating the trial solution: Step 1: Generate the rst guess for the trial solution as follows: Replace all numerical coecients of terms in Q(x) with variable coecients. If there is a sine (or cosine) term, add in the companion cosine (or sine) terms, if they are missing. Then group terms of Q(x) into blocks of terms which are the same up to a power of x, and add in any missing lowerdegree terms in each block. Thus, if a term of the form x n e rx appears in Q(x), ll in the terms of the form e rx [A 0 + A 1 x + + A n x n ], and if a term of the form x n e αx sin(βx) or x n e αx cos(βx) appears in Q(x), ll in the terms of the form e ax cos(bx) [D 0 + D 1 x + + D n x n ] + e ax sin(bx) [E 0 + E 1 x + + E n x n ].
6 Step 2: Solve the homogeneous equation, and write down the general solution. Step 3: Compare the rst guess for the trial solution with the solutions to the homogeneous equation. If any terms overlap, multiply all terms in the overlapping block by the appropriate power of x which will remove the duplication. Here is a series of examples demonstrating the procedure for generating the trial solution: Example: y y = x. Step 1: We ll in the missing constant term in Q(x) to get D 0 + D 1 x. Step 2: General solution is A 1 e x + A 2 e x. Step 3: There is no overlap, so the trial solution is D 0 + D 1 x. Example: y + y = x 2. Step 1: We have D 0 + D 1 x. Step 2: General homogeneous solution is A + Be x. Step 3: There is an overlap (the solution D 0 ) so we multiply the corresponding trial solution terms by x, to get D 0 x + D 1 x 2. Now there is no overlap, so D 0 x + D 1 x 2 Example: y y = e x. is the trial solution. Step 1: We have D 0 e x. Step 2: General homogeneous solution is Ae x + Be x. Step 3: There is an overlap (the solution D 0 e x ) so we multiply the trial solution term by x, to get D 0 xe x. Now there is no overlap, so D 0 xe x is the trial solution. Example: y 2y + y = 3e x. Step 1: We have D 0 e x. Step 2: General homogeneous solution is Ae x + Bxe x. Step 3: There is an overlap (the solution D 0 e x ) so we multiply the trial solution term by x 2, to get rid of the overlap, giving us the trial solution D 0 x 2 e x. Example: y 2y + y = x 3 e x. Step 1: We ll in the lowerdegree terms to get D 0 e x + D 1 xe x + D 2 x 2 e x + D 3 x 3 e x. Step 2: The general homogeneous solution is A 0 e x + A 1 xe x. Step 3: There is an overlap (namely D 0 e x + D 1 xe x ) so we multiply the trial solution terms by x 2 to get D 0 x 2 e x + D 1 x 3 e x + D 2 x 4 e x + D 3 x 5 e x Example: y + y = sin(x). as the trial solution. Step 1: We ll in the missing cosine term to get D 0 cos(x) + E 0 sin(x). Step 2: The general homogeneous solution is A cos(x) + B sin(x). Step 3: There is an overlap (all of D 0 cos(x) + E 0 sin(x)) so we multiply the trial solution terms by x to get D 0 x cos(x) + E 0 x sin(x). There is now no overlap so D 0 x cos(x) + E 0 x sin(x) is the trial solution. Example: y + y = x sin(x). Step 1: We ll in the missing cosine term and then all the lowerdegree terms to get D 0 cos(x) + E 0 sin(x) + D 1 x cos(x) + E 1 x sin(x). Step 2: The general homogeneous solution is A cos(x) + B sin(x). Step 3: There is an overlap (all of D 0 cos(x) + E 0 sin(x)) so we multiply the trial solution terms in that group by x to get D 0 x cos(x) + E 0 x sin(x) + D 1 x 2 cos(x) + E 1 x 2 sin(x), which is the trial solution since now there is no overlap. Example: y y = x + xe x. Step 1: We ll in the lowerdegree term for xe x and the lowerdegree term for x, to get A 0 + A 1 x + B 0 e x + B 1 xe x.
7 Step 2: The general homogeneous solution is C 0 + C 1 x + De x. Step 3: There are overlaps in both groups of terms: A 0 + A 1 x and B 0 e x each overlap, so we multiply the x group by x 2 and the e x group by x to get rid of the overlaps. There are now no additional overlapping terms, so the trial solution is A 0 x 2 + A 1 x 3 + B 0 xe x + B 1 x 2 e x. Example: y + 2y + y = xe x + x cos(x). Step 1: We ll in the lowerdegree term for xe x, then the missing sine term for x cos(x), and then the lowerdegree terms for x cos(x) and x sin(x), to get A 0 e x + A 1 xe x + D 0 cos(x) + E 0 sin(x) + D 1 x cos(x) + E 1 x sin(x). Step 2: The general homogeneous solution is B 0 cos(x) + C 0 sin(x) + B 1 x cos(x) + C 1 x sin(x). Step 3: There is an overlap (namely, all of D 0 cos(x) + E 0 sin(x) + D 1 x cos(x) + D 1 x sin(x)) so we multiply that group by x 2 to get rid of the overlap. There are no additional overlapping terms, so the trial solution is A 0 e x + A 1 xe x + D 0 x 2 cos(x) + E 0 x 2 sin(x) + D 1 x 3 cos(x) + E 1 x 3 sin(x). Here is a series of examples nding the general trial solution and then solving for the coecients: Example: Find a function y such that y + y + y = x. The procedure produces our trial solution as y = D 0 + D 1 x, because there is no overlap with the solutions to the homogeneous equation. We plug in and get 0 + (D 1 ) + (D 1 x + D 0 ) = x, so that D 1 = 1 and D 0 = 1. So our solution is y = x 1. Example: Find a function y such that y y = 2e x. The procedure gives the trial solution as y = D 0 xe x, since D 0 e x overlaps with the solution to the homogeneous equation. If y = D 0 xe x then y = D 0 (x + 2)e x so plugging in yields y y = [D 0 (x + 2)e x ] [D 1 x e x ] = 2e x. Solving yields D 0 = 1, so our solution is y = xe x. Example: Find a function y such that y 2y + y = x + sin(x). The procedure gives the trial solution as y = (D 0 + D 1 x) + (D 2 cos(x) + D 3 sin(x)), by lling in the missing constant term and cosine term, and because there is no overlap with the solutions to the homogeneous equation. Then we have y = D 2 cos(x) D 3 sin(x) and y = D 1 D 2 sin(x) + D 3 cos(x) so plugging in yields y 2y +y = [ D 2 cos(x) D 3 sin(x)] 2 [D 1 D 2 sin(x) + D 3 cos(x)]+[d 0 + D 1 x + D 2 cos(x) + D 3 sin(x)] and setting this equal to x + sin(x) then requires D 0 2D 1 = 0, D 1 = 1, D 2 + 2D 3 D 2 = 1, D 3 2D 2 D 3 = 0, so our solution is y = x cos(x). Example: Find all functions y such that y + y = sin(x). The solutions to the homogeneous system y + y = 0 are y = C 1 cos(x) + C 2 sin(x). Then the procedure gives the trial solution for the nonhomogeneous equation as y = D 0 x cos(x) + D 1 x sin(x), by lling in the missing cosine term and then multiplying both by x due to the overlap with the solutions to the homogeneous equation. We can compute (eventually) that y = D 0 x cos(x) 2D 0 sin(x) D 1 x sin(x) + 2D 1 cos(x). Plugging in yields y +y = ( D 0 x cos(x) 2D 0 sin(x) D 1 x sin(x) + 2D 1 cos(x))+(d 0 x sin(x) + D 1 x cos(x)), and so setting this equal to sin(x), we obtain D 0 = 0 and D 1 = 1 2. Therefore the set of solutions is y = 1 2 x cos(x) + C 1 cos(x) + C 2 sin(x), for constants C 1 and C 2. Advanced Remark: The formal idea behind the method of undetermined coecients is that the terms on the righthand side are themselves solutions of dierential equations and hence are sent to zero by a polynomial in the dierential operator D sending a function to its derivative. (Thus, for example, D(x 2 ) = 2x and D(e x ) = e x.)
8 Then if we apply that polynomial in D which sends Q(x) to zero, to both sides of the original equation, we will end up with a homogeneous equation whose characteristic polynomial is the product of the original characteristic polynomial and the characteristic polynomial for Q(x). Example: Find the form of a solution to y y = x 2. If we dierentiate both sides 3 times (i.e., apply the dierential operator D 3 to both sides), in order to kill o the x 2 term on the righthand side then we get y (5) y (3) = 0, which has characteristic polynomial r 5 r 3 = (r 2 1) (r 3 ). Observe that this polynomial is the product of the characteristic polynomial r 2 1 of the original equation and the polynomial r 3 corresponding to D 3. Now y (5) y (3) = 0 is homogeneous, so we can write down the general solution to obtain y = C 1 + C 2 x + C 3 x 2 + C 4 e x + C 5 e x. This is the same solution form that the method of undetermined coecients gives, though with some extra lowerdegree terms (which are solutions to the homogeneous equation y y = 0). Example: Find the form of a solution to y + y = x + sin(x). This is the same as the equation (D 2 + 1) y = x + sin(x) We want to apply the operator D 2 to kill the x term, and the operator D 2 +1 to kill the sin(x) term. The new dierential equation, after we are done, is (D 2 + 1)(D 2 )(D 2 + 1) y = 0. The characteristic polynomial is (r 2 + 1) 2 r 2, which has a double root at each of r = 0 and ±i. Solving the resulting homogeneous equation gives the solutions as y = C 1 sin(x) + C 2 x sin(x) + C 3 cos(x) + C 4 x cos(x) + C 5 + C 6 x, which is the same thing that the method of undetermined coecients gives, up to some extra terms Variation of Parameters Variation of Parameters will solve any nonhomogeneous linear equation provided that the solutions to the homogeneous equation are known. However, the derivation is not entirely enlightening, so I will just give the steps to follow to solve y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = Q(x). (The method requires being able to solve the homogeneous equation, so we will typically apply this method to the constantcoecient equation y (n) + a n 1 y (n 1) + + a 1 y + a 0 y = 0.) Step 1: Solve the corresponding homogeneous equation y (n) + P n (x) y (n 1) + + P 2 (x) y + P 1 (x) y = 0 and nd n (linearly independent) solutions y 1,, y n. Step 2: Look for functions v 1,, v n making y p = v 1 y 1 + v 2 y v n y n a solution to the original equation: do this by requiring v 1, v 2,, v n to satisfy the system of equations v 1 y 1 + v 2 y v n y n = 0 v 1 y 1 + v 2 y v n y n = 0... v 1 y (n 2) 1 + v 2 y (n 2) v n y n (n 2) = 0 v 1 y (n 1) 1 + v 2 y (n 1) v n y (n 1) n = Q(x) Solve the relations for v 1, v 2,, v n using Cramer's Rule (or any other method). This yields v i = W i(x) W (x), where W is the Wronskian of the functions y 1, y 2,, y n and W i is the same Wronskian determinant except with the ith column replaced with the column vector 0. 0 Q(x) Step 3: Integrate each of these relations to nd v 1,, v n. (Ignore constants of integration.) Step 4: Write down the particular solution to the nonhomogeneous equation, y p = v 1 y 1 + v 2 y v n y n..
9 Step 5: If asked, add the particular solution to the general solution to the homogeneous equation, to nd all solutions of the nonhomogeneous equation. This will yield y = y p + C 1 y C n y n. Plug in any extra conditions given to solve for coecients. Example: Find all functions y for which y + y = sec(x). Step 1: The homogeneous equation is y + y = 0 which has two independent solutions of y 1 = cos(x) and y 2 = sin(x). Step 2: We have Q(x) = sec(x). We have W = cos(x) sin(x) sin(x) cos(x) = 1. Also, W 1 = 0 sin(x) sec(x) cos(x) = sin(x) sec(x). Finally, W 2 = cos(x) 0 sin(x) sec(x) = cos(x) sec(x) = 1. Thus plugging in to the formulas gives v 1 = sin(x) sec(x) = tan(x) and v 2 = cos(x) sec(x) = 1. Step 3: Integrating yields v 1 = ln(cos(x)) and v 2 = x. Step 4: We obtain the particular solution of y p = ln(cos(x)) cos(x) + x sin(x). Step 5: The general solution is, therefore, given by y = [ln(cos(x)) cos(x) + x sin(x)] + C 1 sin(x) + C 2 cos(x). Example: Find all functions y for which y y + y y = e x. We could use undetermined coecients to solve this, but let's use variation of parameters. Step 1: The homogeneous equation is y y + y y = 0 which has characteristic polynomial r 3 r 2 + r 1 = (r 1)(r 2 + 1). So three independent solutions are y 1 = cos(x), y 2 = sin(x), and y 3 = e x. Step 2: We have Q(x) = e x. cos(x) sin(x) e x We have W = sin(x) cos(x) e x R 3+R 1 cos(x) sin(x) e x = sin(x) cos(x) e x cos(x) sin(x) e x 0 0 2e x = 2ex. 0 sin(x) e x Next, W 1 = 0 cos(x) e x e x sin(x) e x = e2x (sin(x) cos(x)). cos(x) 0 e x Also, W 2 = sin(x) 0 e x cos(x) e x e x = e2x (cos(x) + sin(x)). cos(x) sin(x) 0 Finally, W 3 = sin(x) cos(x) 0 cos(x) sin(x) e x = ex. Thus plugging in to the formulas gives v 1 = 1 2 ex (sin(x) cos(x)), v 2 = 1 2 ex (cos(x) + sin(x)), and v 3 = 1 2. Step 3: Integrating yields v 1 = 1 2 ex cos(x), v 2 = 1 2 ex sin(x), and v 3 = 1 2 x. Step 4: The particular solution is y p = 1 2 ex cos 2 (x) 1 2 ex sin 2 (x) xex = 1 2 ex xex. Step 5: The general solution is, therefore, given by y = 1 2 xex + C 1 cos(x) + C 2 sin(x) + C 3 e x. (Note that we absorbed the 1 2 ex term from the particular solution into C 3 e x.) Theorem (Abel's Identity): If y 1, y 2,, y n are any solutions to the homogeneous linear dierential equation y (n) +P n (x) y (n 1) + +P 2 (x) y +P 1 (x) y = 0, then the Wronskian W (y 1,, y n ) is equal to C e P n(x) dx, for some constant C.
10 Abel's Identity allows one to compute the Wronskian (up to a constant factor) without actually nding the solutions y 1, y 2,, y n. Another result of Abel's Identity is that the Wronskian is either zero everywhere (if C = 0) or nonzero everywhere (if C 0, since exponentials are never zero). Abel's Identity provides some relationships between the solution functions, which can sometimes be used to deduce information about them. For example, for a secondorder linear dierential equation with two solutions y 1 and y 2, the Wronskian is W (x) = y 2y 1 y 1y 2. Therefore, if one solution y 1 is known, Abel's identity gives the value of W (x) (up to a value for C, which we can take to be 1 by rescaling), and therefore yields a rstorder linear dierential equation for the other solution y 2, which can then be solved for y 2. This procedure is known in general as reduction of order. The (very clever) proof of Abel's Identity involves showing that the Wronskian W satises the dierential equation W = P n (x) W, by dierentiating the determinant that denes W, and then using row operations to deduce that W = P n (x) W. 4.5 SecondOrder Equations: Applications to Newtonian Mechanics One of the applications we somewhat care about is the use of secondorder dierential equations to solve certain physics problems. Most of the examples involve springs, because springs are easy to talk about. Note: Secondorder linear equations also arise often in basic circuit problems in physics and electrical engineering. All of the discussion of the behaviors of the solutions to these secondorder equations also carries over to that setup Spring Problems and Damping Basic setup: An object is attached to one end of a spring whose other end is xed. The mass is displaced some amount from the equilibrium position, and the problem is to nd the object's position as a function of time. Various modications to this basic setup include any or all of (i) the object slides across a surface thus adding a force (friction) depending on the object's velocity or position, (ii) the object hangs vertically thus adding a constant gravitational force, (iii) a motor or other device imparts some additional nonconstant force (varying with time) to the object. In order to solve problems like this one, follow these steps: Step 1: Draw a diagram and label the quantity or quantities of interest (typically, it is the position of a moving object) and identify and label all forces acting on those quantities. Step 2: Find the values of the forces involved, and then use Newton's Second Law (F = ma) to write down a dierential equation modeling the problem. Also use any information given to write down initial conditions. In the above, F is the net force on the object i.e., the sum of each of the individual forces acting on the mass (with the proper sign) while m is the mass of the object, and a is the object's acceleration. Remember that acceleration is the second derivative of position with respect to time thus, if y(t) is the object's position, acceleration is y (t). You may need to do additional work to solve for unknown constants e.g., for a spring constant, if it is not explicitly given to you before you can fully set up the problem. Step 3: Solve the dierential equation and nd its general solution. Step 4: Plug in any initial conditions to nd the specic solution. Step 5: Check that the answer obtained makes sense in the physical context of the problem.
11 In other words, if you have an object attached to a xed spring sliding on a frictionless surface, you should expect the position to be sinusoidal, something like C 1 sin(ωt) + C 2 cos(ωt) + D for some constants C 1, C 2, ω, D. If you have an object on a spring sliding on a surface imparting friction, you should expect the position to tend to some equilibrium value as t grows to, since the object should be 'slowing down' as time goes on. Basic Example: An object, mass m, is attached to a spring of spring constant k whose other end is xed. The object is displaced a distance d from the equilibrium position of the spring, and is let go with velocity v 0 at time t = 0. If the object is restricted to sliding horizontally on a frictionless surface, nd the position of the object as a function of time. Step 1: Take y(t) to be the displacement of the object from the equilibrium position. The only force acting on the object is from the spring, F spring. Step 2: We know that F spring = k y from Hooke's Law (aka, the only thing we know about springs). Therefore we have the dierential equation k y = m y. We are also given the initial conditions y(0) = d and y (0) = v 0. Step 3: We can rewrite the dierential equation as m y + k y = 0, or as y + k m y = 0. The characteristic equation is then r 2 + k k m = 0 with roots r = ± m i. Hence the general solution is k y = C 1 cos(ωt) + C 2 sin(ωt), where ω = m. Step 4: The initial conditions give d = y(0) = C 1 and v 0 = y (0) = ωc 2 hence C 1 = d and C 2 = v 0 /ω. Hence the solution we want is y = d cos(ωt) + v 0 ω sin(ωt). Step 5: The solution we have obtained makes sense in the context of this problem, since on a frictionless surface we should expect that the object's motion would be purely oscillatory it should just bounce back and forth along the spring forever since there is nothing to slow its motion. We can even see that the k form of the solution agrees with our intuition: the fact that the frequency ω = increases with bigger m spring constant but decreases with bigger mass makes sense a stronger spring with larger k should pull back harder on the object and cause it to oscillate more quickly, while a heavier object should resist the spring's force and oscillate more slowly. Most General Example: An object, mass m, is attached to a spring of spring constant k whose other end is xed. The object is displaced a distance d from the equilibrium position of the spring, and is let go with velocity v 0 at time t = 0. A motor attached to the object imparts a force along its direction of motion given by R(t). If the object is restricted to sliding horizontally on a surface which imparts a frictional force of µ times the velocity of the object (opposite to the object's motion), set up a dierential equation modeling the problem. Here is the diagram:. As before we take y(t) to be the displacement of the object from the equilibrium position. The forces acting on the object are from the spring, F spring, from friction, F friction, and from the motor, F motor.
12 We know that F spring = k y from Hooke's Law (aka, the only thing we know about springs). We are also given that F fric = µ y, since the force acts opposite to the direction of motion and velocity is given by y. And we are just given F motor = R(t). Plugging in gives us the dierential equation k y µ y + R(t) = m y, which in standard form is m y + µ y + k y = R(t). We are also given the initial conditions y(0) = d and y (0) = v 0. Some Terminology: If we were to solve the dierential equation m y + µ y + k y = 0 (here we assume that there is no outside force acting on the object, other than the spring and friction), we would observe a few dierent kinds of behavior depending on the parameters m, µ, and k. Overdamped Case: If µ 2 4mk > 0 and R(t) = 0, we would end up with general solutions of the form C 1 e r1t +C 2 e r2t, which when graphed is just a sum of two exponentiallydecaying functions. Physically, as we can see from the condition µ 2 4mk > 0, this means we have 'too much' friction, since we can just see from the form of the solution function that the position of the object will just slide back towards its equilibrium at y = 0 without oscillating at all. This is the overdamped case. [Overdamped because there is 'too much' damping.] Critically Damped Case: If µ 2 4mk = 0 and R(t) = 0, we would end up with general solutions of the form (C 1 + C 2 t)e rt, which when graphed is a slightlyslowerdecaying exponential function that still does not oscillate, but could possibly cross the position y = 0 once, depending on the values of C 1 and C 2. This is the critically damped case. [Critically because we give the name 'critical' to values where some kind of behavior transitions from one thing to another.] Underdamped Case: If µ 2 4mk < 0 and R(t) = 0, we end up with general solutions of the form e αt [C 1 cos(ωt) + C 2 sin(ωt)], where α = µ 2m and 4mk ω2 µ2 = 4m 2. When graphed this is a sine curve times an exponentiallydecaying function. Physically, this means that there is some friction (the exponential), but 'not enough' friction to eliminate the oscillations entirely the position of the object will still tend toward y = 0, but the sine and cosine terms will ensure that it continues oscillating. This is the underdamped case. [Underdamped because there's not enough damping.] Undamped Case: If there is no friction (i.e., µ = 0), we saw earlier that the solutions are of the form y = C 1 cos(ωt) + C 2 sin(ωt) where ω 2 = k/m. Since there is no friction, it is not a surprise that this is referred to as the undamped case Resonance and Forcing Suppose an object of mass m (sliding on a frictionless surface) is oscillating on a spring with frequency ω. Examine what happens to the object's motion if an external force F (t) = A cos(ωt) is applied which oscillates at the same frequency ω. From the solution to the Basic Example above, we know that ω = k/m so we must have k = m ω 2. Then if y(t) is the position of the object once we add in this new force R(t) = A cos(ωt + θ), Newton's Second Law now gives k y + R(t) = m y, or m y + k y = R(t). If we divide through by m and put in k = m ω 2 we get y + ω 2 y = A m cos(ωt). Now we use the method of undetermined coecients to nd a solution to this dierential equation. We would like to try something of the form y = D 1 cos(ωt) + D 2 sin(ωt), but this will not work because functions of that form are already solutions to the homogeneous equation y + ω 2 y = 0. Instead the method instructs that the appropriate solution will be of the form y = D 1 t cos(ωt) + D 2 t sin(ωt). We can use a trigonometric formula (the sumtoproduct formula) to rewrite this as y = D t cos(ωt + φ), where φ is a phase shift. (We can solve for the coecients in terms of A, m, ω but it will not be so useful.) We can see from this formula that as t grows, so does the amplitude D t: in other words, as time goes on, the object will continue oscillating with frequency ω around its equilibrium point, but the swings back and forth will get larger and larger.
13 You can observe this phenomenon for yourself if you sit in a rocking chair, or swing an object back and forth you will quickly nd that the most eective way to rock the chair or swing the object is to push back and forth at the same frequency that the object is already moving at. We may work out the same computation with an external force F (t) = A cos(ω 1 t) oscillating at a frequency ω 1 ω. In this case (using the same argument as above) we have y + ω 2 y = A m cos(ω 1t). The trial solution (again by undetermined coecients) is y(t) = B cos(ω 1 t), where B = A/m ω 2 ω1 2. Thus the overall solution is B cos(ω 1 t), plus a solution to the homogeneous system. Now as we can see, if ω 1 and ω are far apart (i.e., the driving force is oscillating at a very dierent frequency from the frequency of the original system) then B will be small, and so the overall change B cos(ω 1 t) that the driving force adds will be relatively small. However, if ω 1 and ω are very close to one another (i.e., the driving force is oscillating at a frequency close to that of the original system) then B will be large, and so the driving force will cause the system to oscillate with a much bigger amplitude. As ω 1 approaches ω, the amplitude B will go to, which agrees with the behavior seen in the previous example (where we took ω 1 = ω). Important Remark: Understanding how resonance arises (and how to minimize it!) is a very, very important application of dierential equations to structural engineering. A poor understanding of resonance is something which, at several times in the nottoodistant past, has caused bridges to fall down, airplanes to crash, and buildings to fall over. We can see from the two examples that resonance arises when an external force acts on a system at (or very close to) the same frequency that the system is already oscillating at. Of course, resonance is not always bad. The general principle, of applying an external driving force at (one of) a system's natural resonance frequencies, is the underlying physical idea behind the construction of many types of musical instruments. Well, you're at the end of my handout. Hope it was helpful. Copyright notice: This material is copyright Evan Dummit, You may not reproduce or distribute this material without my express permission.
EXAM. Practice Questions for Exam #2. Math 3350, Spring April 3, 2004 ANSWERS
EXAM Practice Questions for Exam #2 Math 3350, Spring 2004 April 3, 2004 ANSWERS i Problem 1. Find the general solution. A. D 3 (D 2)(D 3) 2 y = 0. The characteristic polynomial is λ 3 (λ 2)(λ 3) 2. Thus,
More informationHigher Order Linear Differential Equations with Constant Coefficients
Higher Order Linear Differential Equations with Constant Coefficients Part I. Homogeneous Equations: Characteristic Roots Objectives: Solve nth order homogeneous linear equations where a n,, a 1, a 0
More informationSecond Order Linear Differential Equations
CHAPTER 2 Second Order Linear Differential Equations 2.. Homogeneous Equations A differential equation is a relation involving variables x y y y. A solution is a function f x such that the substitution
More informationChapter 2 Second Order Differential Equations
Chapter 2 Second Order Differential Equations Either mathematics is too big for the human mind or the human mind is more than a machine.  Kurt Gödel (19061978) 2.1 The Simple Harmonic Oscillator The
More informationApplications of SecondOrder Differential Equations
Applications of SecondOrder Differential Equations Secondorder linear differential equations have a variety of applications in science and engineering. In this section we explore two of them: the vibration
More informationSECONDORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS
L SECONDORDER LINEAR HOOGENEOUS DIFFERENTIAL EQUATIONS SECONDORDER LINEAR HOOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS A secondorder linear differential equation is one of the form d
More informationPhysics 53. Oscillations. You've got to be very careful if you don't know where you're going, because you might not get there.
Physics 53 Oscillations You've got to be very careful if you don't know where you're going, because you might not get there. Yogi Berra Overview Many natural phenomena exhibit motion in which particles
More informationNonhomogeneous Linear Equations
Nonhomogeneous Linear Equations In this section we learn how to solve secondorder nonhomogeneous linear differential equations with constant coefficients, that is, equations of the form ay by cy G x where
More informationSecondOrder Linear Differential Equations
SecondOrder Linear Differential Equations A secondorder linear differential equation has the form 1 Px d 2 y dx 2 dy Qx dx Rxy Gx where P, Q, R, and G are continuous functions. We saw in Section 7.1
More informationHomework #7 Solutions
MAT 0 Spring 201 Problems Homework #7 Solutions Section.: 4, 18, 22, 24, 4, 40 Section.4: 4, abc, 16, 18, 22. Omit the graphing part on problems 16 and 18...4. Find the general solution to the differential
More informationSolving the Harmonic Oscillator Equation. Morgan Root NCSU Department of Math
Solving the Harmonic Oscillator Equation Morgan Root NCSU Department of Math SpringMass System Consider a mass attached to a wall by means of a spring. Define y to be the equilibrium position of the block.
More informationA Brief Review of Elementary Ordinary Differential Equations
1 A Brief Review of Elementary Ordinary Differential Equations At various points in the material we will be covering, we will need to recall and use material normally covered in an elementary course on
More informationLecture 31: Second order homogeneous equations II
Lecture 31: Second order homogeneous equations II Nathan Pflueger 21 November 2011 1 Introduction This lecture gives a complete description of all the solutions to any differential equation of the form
More informationMechanical Vibrations
Mechanical Vibrations A mass m is suspended at the end of a spring, its weight stretches the spring by a length L to reach a static state (the equilibrium position of the system). Let u(t) denote the displacement,
More information2 Integrating Both Sides
2 Integrating Both Sides So far, the only general method we have for solving differential equations involves equations of the form y = f(x), where f(x) is any function of x. The solution to such an equation
More informationHooke s Law. Spring. Simple Harmonic Motion. Energy. 12/9/09 Physics 201, UWMadison 1
Hooke s Law Spring Simple Harmonic Motion Energy 12/9/09 Physics 201, UWMadison 1 relaxed position F X = kx > 0 F X = 0 x apple 0 x=0 x > 0 x=0 F X =  kx < 0 x 12/9/09 Physics 201, UWMadison 2 We know
More informationSecond Order Linear Nonhomogeneous Differential Equations; Method of Undetermined Coefficients. y + p(t) y + q(t) y = g(t), g(t) 0.
Second Order Linear Nonhomogeneous Differential Equations; Method of Undetermined Coefficients We will now turn our attention to nonhomogeneous second order linear equations, equations with the standard
More information1 of 10 11/23/2009 6:37 PM
hapter 14 Homework Due: 9:00am on Thursday November 19 2009 Note: To understand how points are awarded read your instructor's Grading Policy. [Return to Standard Assignment View] Good Vibes: Introduction
More informationPeriodic Motion or Oscillations. Physics 232 Lecture 01 1
Periodic Motion or Oscillations Physics 3 Lecture 01 1 Periodic Motion Periodic Motion is motion that repeats about a point of stable equilibrium Stable Equilibrium Unstable Equilibrium A necessary requirement
More informationGeneral Theory of Differential Equations Sections 2.8, 3.13.2, 4.1
A B I L E N E C H R I S T I A N U N I V E R S I T Y Department of Mathematics General Theory of Differential Equations Sections 2.8, 3.13.2, 4.1 Dr. John Ehrke Department of Mathematics Fall 2012 Questions
More informationSo far, we have looked at homogeneous equations
Chapter 3.6: equations Nonhomogeneous So far, we have looked at homogeneous equations L[y] = y + p(t)y + q(t)y = 0. Homogeneous means that the right side is zero. Linear homogeneous equations satisfy
More informationContents. 1 FirstOrder Dierential Equations. 1.1 Introduction and Terminology
Dierential Equations (part 1) : FirstOrder Dierential Equations (by Evan Dummit, 2016, v. 2.10) Contents 1 FirstOrder Dierential Equations 1 1.1 Introduction and Terminology........................................
More informationFinding y p in ConstantCoefficient Nonhomogenous Linear DEs
Finding y p in ConstantCoefficient Nonhomogenous Linear DEs Introduction and procedure When solving DEs of the form the solution looks like ay + by + cy =, y = y c + y p where y c is the complementary
More information3.2 Sources, Sinks, Saddles, and Spirals
3.2. Sources, Sinks, Saddles, and Spirals 6 3.2 Sources, Sinks, Saddles, and Spirals The pictures in this section show solutions to Ay 00 C By 0 C Cy D 0. These are linear equations with constant coefficients
More informationtegrals as General & Particular Solutions
tegrals as General & Particular Solutions dy dx = f(x) General Solution: y(x) = f(x) dx + C Particular Solution: dy dx = f(x), y(x 0) = y 0 Examples: 1) dy dx = (x 2)2 ;y(2) = 1; 2) dy ;y(0) = 0; 3) dx
More informationLectures 56: Taylor Series
Math 1d Instructor: Padraic Bartlett Lectures 5: Taylor Series Weeks 5 Caltech 213 1 Taylor Polynomials and Series As we saw in week 4, power series are remarkably nice objects to work with. In particular,
More informationSystems of Linear Equations in Two Variables
Chapter 6 Systems of Linear Equations in Two Variables Up to this point, when solving equations, we have always solved one equation for an answer. However, in the previous chapter, we saw that the equation
More informationNotes from FE Review for mathematics, University of Kentucky
Notes from FE Review for mathematics, University of Kentucky Dr. Alan Demlow March 19, 2012 Introduction These notes are based on reviews for the mathematics portion of the Fundamentals of Engineering
More informationDynamics. Figure 1: Dynamics used to generate an exemplar of the letter A. To generate
Dynamics Any physical system, such as neurons or muscles, will not respond instantaneously in time but will have a timevarying response termed the dynamics. The dynamics of neurons are an inevitable constraint
More informationExperiment Type: OpenEnded
Simple Harmonic Oscillation Overview Experiment Type: OpenEnded In this experiment, students will look at three kinds of oscillators and determine whether or not they can be approximated as simple harmonic
More informationAP1 Oscillations. 1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false?
1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false? (A) The displacement is directly related to the acceleration. (B) The
More information1 of 7 10/2/2009 1:13 PM
1 of 7 10/2/2009 1:13 PM Chapter 6 Homework Due: 9:00am on Monday, September 28, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View]
More informationSimple Harmonic Motion
Simple Harmonic Motion Simple harmonic motion is one of the most common motions found in nature and can be observed from the microscopic vibration of atoms in a solid to rocking of a supertanker on the
More informationLecture L19  Vibration, Normal Modes, Natural Frequencies, Instability
S. Widnall 16.07 Dynamics Fall 2009 Version 1.0 Lecture L19  Vibration, Normal Modes, Natural Frequencies, Instability Vibration, Instability An important class of problems in dynamics concerns the free
More informationTaylor Polynomials and Taylor Series Math 126
Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will
More informationLecture 6. Weight. Tension. Normal Force. Static Friction. Cutnell+Johnson: 4.84.12, second half of section 4.7
Lecture 6 Weight Tension Normal Force Static Friction Cutnell+Johnson: 4.84.12, second half of section 4.7 In this lecture, I m going to discuss four different kinds of forces: weight, tension, the normal
More informationSimple Harmonic Motion
Simple Harmonic Motion Theory Simple harmonic motion refers to the periodic sinusoidal oscillation of an object or quantity. Simple harmonic motion is eecuted by any quantity obeying the Differential
More informationA Second Course in Elementary Differential Equations: Problems and Solutions. Marcel B. Finan Arkansas Tech University c All Rights Reserved
A Second Course in Elementary Differential Equations: Problems and Solutions Marcel B. Finan Arkansas Tech University c All Rights Reserved Contents 8 Calculus of MatrixValued Functions of a Real Variable
More informationSection 1.1. Introduction to R n
The Calculus of Functions of Several Variables Section. Introduction to R n Calculus is the study of functional relationships and how related quantities change with each other. In your first exposure to
More information19.6. Finding a Particular Integral. Introduction. Prerequisites. Learning Outcomes. Learning Style
Finding a Particular Integral 19.6 Introduction We stated in Block 19.5 that the general solution of an inhomogeneous equation is the sum of the complementary function and a particular integral. We have
More informationSimple Harmonic Motion Concepts
Simple Harmonic Motion Concepts INTRODUCTION Have you ever wondered why a grandfather clock keeps accurate time? The motion of the pendulum is a particular kind of repetitive or periodic motion called
More informationcorrectchoice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were:
Topic 1 2.1 mode MultipleSelection text How can we approximate the slope of the tangent line to f(x) at a point x = a? This is a Multiple selection question, so you need to check all of the answers that
More information1 Introduction to Differential Equations
1 Introduction to Differential Equations A differential equation is an equation that involves the derivative of some unknown function. For example, consider the equation f (x) = 4x 3. (1) This equation
More information3.2 The Factor Theorem and The Remainder Theorem
3. The Factor Theorem and The Remainder Theorem 57 3. The Factor Theorem and The Remainder Theorem Suppose we wish to find the zeros of f(x) = x 3 + 4x 5x 4. Setting f(x) = 0 results in the polynomial
More informationr (t) = 2r(t) + sin t θ (t) = r(t) θ(t) + 1 = 1 1 θ(t) 1 9.4.4 Write the given system in matrix form x = Ax + f ( ) sin(t) x y 1 0 5 z = dy cos(t)
Solutions HW 9.4.2 Write the given system in matrix form x = Ax + f r (t) = 2r(t) + sin t θ (t) = r(t) θ(t) + We write this as ( ) r (t) θ (t) = ( ) ( ) 2 r(t) θ(t) + ( ) sin(t) 9.4.4 Write the given system
More informationPolynomial Degree and Finite Differences
CONDENSED LESSON 7.1 Polynomial Degree and Finite Differences In this lesson you will learn the terminology associated with polynomials use the finite differences method to determine the degree of a polynomial
More informationLABORATORY 9. Simple Harmonic Motion
LABORATORY 9 Simple Harmonic Motion Purpose In this experiment we will investigate two examples of simple harmonic motion: the massspring system and the simple pendulum. For the massspring system we
More informationORDINARY DIFFERENTIAL EQUATIONS
Page 1 of 21 ORDINARY DIFFERENTIAL EQUATIONS Lecture 15 Physical Systems Modelled by Linear Homogeneous Second Order ODEs (Revised 22 March, 2009 @ 18:00) Professor Stephen H Saperstone Department of
More informationSimple Harmonic Motion
Simple Harmonic Motion 1 Object To determine the period of motion of objects that are executing simple harmonic motion and to check the theoretical prediction of such periods. 2 Apparatus Assorted weights
More informationSECONDORDER LINEAR DIFFERENTIAL EQUATIONS
SECONDORDER LINEAR DIFFERENTIAL EQUATIONS A secondorder linear differential equation has the form 1 Px d y dx dy Qx dx Rxy Gx where P, Q, R, and G are continuous functions. Equations of this type arise
More informationLab M1: The Simple Pendulum
Lab M1: The Simple Pendulum Introduction. The simple pendulum is a favorite introductory exercise because Galileo's experiments on pendulums in the early 1600s are usually regarded as the beginning of
More informationAn important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate.
Chapter 10 Series and Approximations An important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate 1 0 e x2 dx, you could set
More informationChapter 1. Oscillations. Oscillations
Oscillations 1. A mass m hanging on a spring with a spring constant k has simple harmonic motion with a period T. If the mass is doubled to 2m, the period of oscillation A) increases by a factor of 2.
More informationSimple Harmonic Motion
5 Simple Harmonic Motion Note: this section is not part of the syllabus for PHYS26. You should already be familiar with simple harmonic motion from your first year course PH115 Oscillations and Waves.
More information19.7. Applications of Differential Equations. Introduction. Prerequisites. Learning Outcomes. Learning Style
Applications of Differential Equations 19.7 Introduction Blocks 19.2 to 19.6 have introduced several techniques for solving commonlyoccurring firstorder and secondorder ordinary differential equations.
More information3.3. Solving Polynomial Equations. Introduction. Prerequisites. Learning Outcomes
Solving Polynomial Equations 3.3 Introduction Linear and quadratic equations, dealt within Sections 3.1 and 3.2, are members of a class of equations, called polynomial equations. These have the general
More informationMATH Fundamental Mathematics II.
MATH 10032 Fundamental Mathematics II http://www.math.kent.edu/ebooks/10032/funmath2.pdf Department of Mathematical Sciences Kent State University December 29, 2008 2 Contents 1 Fundamental Mathematics
More informationUpdated 2013 (Mathematica Version) M1.1. Lab M1: The Simple Pendulum
Updated 2013 (Mathematica Version) M1.1 Introduction. Lab M1: The Simple Pendulum The simple pendulum is a favorite introductory exercise because Galileo's experiments on pendulums in the early 1600s are
More informationZeros of Polynomial Functions
Review: Synthetic Division Find (x 25x  5x 3 + x 4 ) (5 + x). Factor Theorem Solve 2x 35x 2 + x + 2 =0 given that 2 is a zero of f(x) = 2x 35x 2 + x + 2. Zeros of Polynomial Functions Introduction
More informationREVIEW EXERCISES DAVID J LOWRY
REVIEW EXERCISES DAVID J LOWRY Contents 1. Introduction 1 2. Elementary Functions 1 2.1. Factoring and Solving Quadratics 1 2.2. Polynomial Inequalities 3 2.3. Rational Functions 4 2.4. Exponentials and
More informationDecember 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B. KITCHENS
December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B KITCHENS The equation 1 Lines in twodimensional space (1) 2x y = 3 describes a line in twodimensional space The coefficients of x and y in the equation
More informationSouth Carolina College and CareerReady (SCCCR) PreCalculus
South Carolina College and CareerReady (SCCCR) PreCalculus Key Concepts Arithmetic with Polynomials and Rational Expressions PC.AAPR.2 PC.AAPR.3 PC.AAPR.4 PC.AAPR.5 PC.AAPR.6 PC.AAPR.7 Standards Know
More informationThnkwell s Homeschool Precalculus Course Lesson Plan: 36 weeks
Thnkwell s Homeschool Precalculus Course Lesson Plan: 36 weeks Welcome to Thinkwell s Homeschool Precalculus! We re thrilled that you ve decided to make us part of your homeschool curriculum. This lesson
More informationCopyrighted Material. Chapter 1 DEGREE OF A CURVE
Chapter 1 DEGREE OF A CURVE Road Map The idea of degree is a fundamental concept, which will take us several chapters to explore in depth. We begin by explaining what an algebraic curve is, and offer two
More informationMathematics Course 111: Algebra I Part IV: Vector Spaces
Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 19967 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are
More informationSolving Systems of Linear Equations. Substitution
Solving Systems of Linear Equations There are two basic methods we will use to solve systems of linear equations: Substitution Elimination We will describe each for a system of two equations in two unknowns,
More informationNewton s Laws of Motion
Section 3.2 Newton s Laws of Motion Objectives Analyze relationships between forces and motion Calculate the effects of forces on objects Identify force pairs between objects New Vocabulary Newton s first
More information5 Indefinite integral
5 Indefinite integral The most of the mathematical operations have inverse operations: the inverse operation of addition is subtraction, the inverse operation of multiplication is division, the inverse
More informationRevised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m)
Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations.
More informationSystem of First Order Differential Equations
CHAPTER System of First Order Differential Equations In this chapter, we will discuss system of first order differential equations. There are many applications that involving find several unknown functions
More informationSTUDY GUIDE FOR SOME BASIC INTERMEDIATE ALGEBRA SKILLS
STUDY GUIDE FOR SOME BASIC INTERMEDIATE ALGEBRA SKILLS The intermediate algebra skills illustrated here will be used extensively and regularly throughout the semester Thus, mastering these skills is an
More informationSample Solutions of Assignment 5 for MAT3270B: Section 3.4. In each of problems find the general solution of the given differential equation
Sample Solutions of Assignment 5 for MAT370B: 3.43.9 Section 3.4 In each of problems find the general solution of the given differential equation 7. y y + y = 0 1. 4y + 9y = 0 14. 9y + 9y 4y = 0 15. y
More informationComputer Experiment. Simple Harmonic Motion. Kinematics and Dynamics of Simple Harmonic Motion. Evaluation copy
INTRODUCTION Simple Harmonic Motion Kinematics and Dynamics of Simple Harmonic Motion Computer Experiment 16 When you suspend an object from a spring, the spring will stretch. If you pull on the object,
More informationFourier Series. A Fourier series is an infinite series of the form. a + b n cos(nωx) +
Fourier Series A Fourier series is an infinite series of the form a b n cos(nωx) c n sin(nωx). Virtually any periodic function that arises in applications can be represented as the sum of a Fourier series.
More informationZeros of Polynomial Functions
Zeros of Polynomial Functions The Rational Zero Theorem If f (x) = a n x n + a n1 x n1 + + a 1 x + a 0 has integer coefficients and p/q (where p/q is reduced) is a rational zero, then p is a factor of
More informationDeterminants can be used to solve a linear system of equations using Cramer s Rule.
2.6.2 Cramer s Rule Determinants can be used to solve a linear system of equations using Cramer s Rule. Cramer s Rule for Two Equations in Two Variables Given the system This system has the unique solution
More informationMATH10212 Linear Algebra. Systems of Linear Equations. Definition. An ndimensional vector is a row or a column of n numbers (or letters): a 1.
MATH10212 Linear Algebra Textbook: D. Poole, Linear Algebra: A Modern Introduction. Thompson, 2006. ISBN 0534405967. Systems of Linear Equations Definition. An ndimensional vector is a row or a column
More information9.2 Summation Notation
9. Summation Notation 66 9. Summation Notation In the previous section, we introduced sequences and now we shall present notation and theorems concerning the sum of terms of a sequence. We begin with a
More informationFourier Series, Integrals, and Transforms
Chap. Sec.. Fourier Series, Integrals, and Transforms Fourier Series Content: Fourier series (5) and their coefficients (6) Calculation of Fourier coefficients by integration (Example ) Reason hy (6) gives
More informationZeros of a Polynomial Function
Zeros of a Polynomial Function An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we
More information2.2 Magic with complex exponentials
2.2. MAGIC WITH COMPLEX EXPONENTIALS 97 2.2 Magic with complex exponentials We don t really know what aspects of complex variables you learned about in high school, so the goal here is to start more or
More informationASEN 3112  Structures. MDOF Dynamic Systems. ASEN 3112 Lecture 1 Slide 1
19 MDOF Dynamic Systems ASEN 3112 Lecture 1 Slide 1 A TwoDOF MassSpringDashpot Dynamic System Consider the lumpedparameter, massspringdashpot dynamic system shown in the Figure. It has two point
More informationA C O U S T I C S of W O O D Lecture 3
Jan Tippner, Dep. of Wood Science, FFWT MU Brno jan. tippner@mendelu. cz Content of lecture 3: 1. Damping 2. Internal friction in the wood Content of lecture 3: 1. Damping 2. Internal friction in the wood
More informationEigenvalues and Eigenvectors
Chapter 6 Eigenvalues and Eigenvectors 6. Introduction to Eigenvalues Linear equations Ax D b come from steady state problems. Eigenvalues have their greatest importance in dynamic problems. The solution
More informationA) F = k x B) F = k C) F = x k D) F = x + k E) None of these.
CT161 Which of the following is necessary to make an object oscillate? i. a stable equilibrium ii. little or no friction iii. a disturbance A: i only B: ii only C: iii only D: i and iii E: All three Answer:
More informationWEEK #5: Trig Functions, Optimization
WEEK #5: Trig Functions, Optimization Goals: Trigonometric functions and their derivatives Optimization Textbook reading for Week #5: Read Sections 1.8, 2.10, 3.3 Trigonometric Functions From Section 1.8
More informationSome Notes on Taylor Polynomials and Taylor Series
Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited
More informationModule M6.3 Solving second order differential equations
F L E X I B L E L E A R N I N G A P P R O A C H T O P H Y S I C S Module M6. Solving second order differential equations Opening items. Module introduction.2 Fast track questions. Rea to stu? 2 Methods
More information{ f (x)e sx if s = r. f(x)e sx if s r. (D r) n y = e rx f(x),
1 Recipes for Solving ConstantCoefficient Linear Nonhomogeneous Ordinary Differential Equations Whose Right Hand Sides are Sums of Products of Polynomials times Exponentials Let D = d dx. We begin by
More informationMath 2280 Section 002 [SPRING 2013] 1
Math 2280 Section 002 [SPRING 2013] 1 Today well learn about a method for solving systems of differential equations, the method of elimination, that is very similar to the elimination methods we learned
More informationSOLVING POLYNOMIAL EQUATIONS
C SOLVING POLYNOMIAL EQUATIONS We will assume in this appendix that you know how to divide polynomials using long division and synthetic division. If you need to review those techniques, refer to an algebra
More informationSolving Second Order Linear ODEs. Table of contents
Solving Second Order Linear ODEs Table of contents Solving Second Order Linear ODEs........................................ 1 1. Introduction: A few facts............................................. 1.
More informationPURSUITS IN MATHEMATICS often produce elementary functions as solutions that need to be
Fast Approximation of the Tangent, Hyperbolic Tangent, Exponential and Logarithmic Functions 2007 Ron Doerfler http://www.myreckonings.com June 27, 2007 Abstract There are some of us who enjoy using our
More informationWEEK 6: FORCE, MASS, AND ACCELERATION
Name Date Partners WEEK 6: FORCE, MASS, AND ACCELERATION OBJECTIVES To develop a definition of mass in terms of an object s acceleration under the influence of a force. To find a mathematical relationship
More informationThe Method of Partial Fractions Math 121 Calculus II Spring 2015
Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method
More informationCHAPTER 2 FOURIER SERIES
CHAPTER 2 FOURIER SERIES PERIODIC FUNCTIONS A function is said to have a period T if for all x,, where T is a positive constant. The least value of T>0 is called the period of. EXAMPLES We know that =
More informationVELOCITY, ACCELERATION, FORCE
VELOCITY, ACCELERATION, FORCE velocity Velocity v is a vector, with units of meters per second ( m s ). Velocity indicates the rate of change of the object s position ( r ); i.e., velocity tells you how
More informationBasic Terminology for Systems of Equations in a Nutshell. E. L. Lady. 3x 1 7x 2 +4x 3 =0 5x 1 +8x 2 12x 3 =0.
Basic Terminology for Systems of Equations in a Nutshell E L Lady A system of linear equations is something like the following: x 7x +4x =0 5x +8x x = Note that the number of equations is not required
More informationUp and Down: Damped Harmonic Motion
Up and Down: Damped Harmonic Motion Activity 27 An object hanging from a spring can bounce up and down in a simple way. The vertical position of the object can be described mathematically in terms of a
More informationRecall that two vectors in are perpendicular or orthogonal provided that their dot
Orthogonal Complements and Projections Recall that two vectors in are perpendicular or orthogonal provided that their dot product vanishes That is, if and only if Example 1 The vectors in are orthogonal
More information