Ordinary Differential Equations

Transcription

1 Ordinary Differential Equations Henry Liu, 22 December 2008 An ordinary differential equation (ODE) is an equation which relates two variables, say, x and y, and derivatives, d2 y, d3 y,.... If dn y is highest ordered derivative, then the 2 3 n ODE has n th order. To solve an ODE means that we would like to find an equation linking y and x, without derivatives. Whenever possible, one should try and get y explicitly in terms of x, although this is sometimes not easy. If we are asked for a general solution, then we have to determine the form of all possible solutions, and we will have arbitrary constants (such as A, B, C,... ) involved. If we are given more information (such as y =? when x =? ), then we can determine the solution more specifically, by determining the arbitrary constant(s) in the general solution. Such a solution is called a particular solution. We shall discuss five types of ODEs. Three of these types will be of first order, and the other two types will be of second order.. First Order, Variable Separable ODEs If it is possible to rearrange a given first ordered ODE so that terms involving x are on one side, and terms involving y are on the other side, then the ODE is variable separable. We may then solve such an ODE by integrating both sides, after rearranging. Example. Find the general solution to = e3y sin x. Also, find the particular solution where y = 0 when x = π. Solution. The given ODE can be rearranged to e 3y = sin x. Integrating both sides, we have e 3y = sin x 3 e 3y = cos x + C e 3y = 3 cos x + D where D = 3C 3y = ln(3 cos x + D) y = ln(3 cos x + D), 3 which is the general solution.

2 For the particular solution, we substitute y = 0 and x = π to find D. We have 0 = ln(3 cos π + D) 3 0 = ln( 3 + D) = 3 + D 4 = D. The particular solution is y = ln(3 cos x + 4) Solving First Order ODEs with the Integrating Factor Method If we have an ODE in the form + yp (x) = Q(x), () where P (x) and Q(x) are given function of x, then we may solve the ODE () as follows. Let R(x) = e P (x), where we ignore the arbitrary constant in the integral. Note that dr = e P (x) P (x) = R(x)P (x). Now, observe that, by the product rule, we have d dr [yr(x)] = R(x) + y d [yr(x)] = R(x) + yp (x)r(x). (2) So, if we multiply each term of () by R(x), and then applying (2), we find that R(x) + yp (x)r(x) = Q(x)R(x) d [yr(x)] = Q(x)R(x). (3) Hence, if we integrate both sides of (3) with respect to x, and then divide by R(x), we will have solved the ODE (). The function R(x) = e P (x) is called the integrating factor (IF), and this method is known as the integrating factor method. So, to summarise in a more compact form, if we are given an ODE and we can rearrange it to the form + yp (x) = Q(x), then the procedure to solving it is as follows. Find the integrating factor (IF) R(x) = e P (x) (ignore the constant of integration). Then, write down the equation d [yr(x)] = Q(x)R(x). 2

3 Finally, integrate both sides with respect to x, and then divide by R(x). Example 2. Find the general solution to x 2y = x3 +. Also, find the particular solution where y() = 3 2. Solution. The given ODE can be rearranged to So, the IF is 2 x y = x2 + x. R(x) = e 2 x = e 2 ln x = (e ln x ) 2 = x 2 = x 2. So, we have d ( [yx 2 ] = x 2 + ) x 2 = + x 3 x yx 2 = ( + x 3 ) = x 2 x 2 + C y = x Cx2, which is the general solution. For the particular solution, we substitute y = = C 2 C =. The particular solution is y = x x2. 3. Homogeneous First Order ODEs A first ordered ODE is called homogeneous if it is of the form ( y ) = f x for some function f. For example, the ODE (2x y) = 2y x. is homogeneous, since it can be written as = 2( y ) x 2 y. x 3 and x = to find C. We have

4 We can solve a homogeneous ODE as follows. Making the substitution y = vx and using the product rule, we have f(v) = = xdv + v, which is now a variable separable ODE between v and x. Indeed, we have f(v) v dv = + C. x We can then find v in terms of x, and substituting v = y back into the solution then x gives y in terms of x. Example 3. Find the general solution to (2x y) = 2y x. Solution. As we have alrea seen, we can re-write the ODE as = 2( y ) x 2 y. x so that it is homogeneous. So, let y = vx, so that = x dv + v. Hence, x dv + v = 2v 2 v x dv = 2v 2 v 2 v v 2 dv = x v = 2v v(2 v) 2 v = v2 2 v We can find the integral on the left by the usual partial fractions method: write 2 v v 2 = A v + B v +, and solve for A and B. We find that A = and B = 3. So, v v 2 dv = 2 v dv 3 2 v + dv 2 ln v 3 ln v + = 2 ln = 2 ln v 3 ln v + 2 = ln x + C x v v + 3 = 2 ln x + 2C = ln(dx2 ), where 2C = ln D v v + 3 = Dx2. 4

5 So, replacing v with y x gives y x y x + 3 = Dx2, or y x = Dx 2 y x + 3. We cannot really simplify much further, but if we were given that y x, then the two terms inside the modulus signs are non-negative, and we can ti up slightly to y x = D(y + x) 3. There are some further examples of ODEs which can be solved by this method, if one makes a change of coordinates. Example 4. Find the general solution to the ODE = x + y 3x y + 5. Solution. For this, we cannot quite use the previous method directly. We consider the simultaneous equations x + y = 0, 3x y + 5 = 0. These represent two lines in the (x, y)-plane which are not parallel, and hence they meet at a point. Solving the system, we easily find that x = and y = 2. So, we make the substitution X = x + and Y = y 2. This gives us dy dx = = x + y 3x y + 5 (X ) + (Y + 2) = 3(X ) (Y + 2) + 5 = X + Y 3X Y = + Y X. 3 Y X From here, we can proceed as before. Letting Y = vx, we again have X dv dx = + v 3 v 3 v (v ) 2 dv = X dx. v = + v v(3 v) 3 v = 2v + v2 3 v = (v )2 3 v Again, we must split 3 v (v ) 2 into partial fractions. Writing 3 v (v ) = A 2 v + B (v ) 2 and solving for A and B, we have A = and B = 2. So, 3 v (v ) dv = 2 v dv + 2 dv = ln v 2(v ) (v ) 2 5

6 ln v 2(v ) = dx = ln X + C X 2(v ) = ln X + ln D + ln v = ln(d X v ) = ln(d Xv X ), where C = ln D 2 = ln(d Y X ) Y X 2X = ln(d Y X ) Y X 2(x + ) = ln(d (y 2) (x + ) ) (y 2) (x + ) 2(x + ) = ln(d y x 3 ), y x 3 which is very much as simple as possible. What if the simultaneous equations have no solutions? That is, what if the corresponding lines are parallel? This turns out to be fairly straightforward. We just make a slightly different substitution. Example 5. Find the general solution to the ODE = 3x + 2y + 3x + 2y + 6. Solution. The lines 3x + 2y + = 0 and 3x + 2y + 6 = 0 are parallel, so we cannot use the same method as in the previous example. But, we can just substitute z for either 3x+2y + or 3x+2y +6. Say, z = 3x+2y +. Then, dz = 3+2, and then substituting the given ODE into this will give us a variable separable equation. We have, dz + 2y + = x 3x + 2y + 6 = 3 + 2z z + 5 z + 5 5z + 5 dz = = x + C. We can write so that = 3(z + 5) + 2z z + 5 z + 5 5z + 5 = ( z + 5 ) = ( + 2 ), 5 z z + 3 ( ) dz = x + C z + 3 (z + 2 ln z + 3 ) 5 = x + C z + ln[(z + 3) 2 ] = 5x + D, where D = 5C 3x + 2y + + ln[(3x + 2y + + 3) 2 ] = 5x + D = 5z + 5 z + 5 2x + 2y + ln[(3x + 2y + 4) 2 ] = E, where E = D. 6

7 4. Second Order ODEs with Constant Coefficients We shall describe the method of how to solve an ODE of the form a d2 y + b + cy = f(x), (4) 2 where a, b and c are given, real constants, with a 0, and f(x) is a given function of x, called the forcing function of the ODE. We first consider the case f(x) 0. That is, a d2 y + b + cy = 0. (5) 2 In this case, the ODE is said to be homogeneous. To solve (5), we write out the auxiliary equation aλ 2 + bλ + c = 0, (6) and solve it (either by factorisation, or by the quadratic formula). There are then three possible forms of the general solution to (5), depending on the solutions of (6) If the solutions of (6) are two distinct real roots λ and λ 2, then the general solution to (5) is y = Ae λ x + Be λ 2x. If the solution of (6) is one (repeated) real root λ, then the general solution to (5) is y = (Ax + B)e λ x. If the solutions of (6) are two complex conjugate roots p + iq and p iq, then the general solution to (5) is y = e px (A sin qx + B cos qx). In each case, A and B are arbitrary constants (as usual). Example 6. The general solutions to are, respectively 2 3y = 0, 2 9d2 y y = 0, y = 0 y = Ae 3x + Be x, y = (Ax + B)e x/3, y = e 2x (A sin 3x + B cos 3x). Now, we go back to the ODE (4). The solution will be made up of two parts. One part, called the complementary function (CF), is obtained by solving the corresponding ODE (5). The other part, called the particular integral (PI), is obtained by looking at the function f(x) in (4). The PI takes a form rather similar to f(x), and we will have to 7

8 determine some unknown constants to find it. We shall write y c and for the CF and PI respectively. The final solution to (4) is then y = y c +. The most well-known examples of f(x) are when it involves a polynomial, an exponential, a trig function: sine or cosine, or a combination of these. We now look at plenty of examples. Example 7. Find the general solution of y = 2e3x. Solution. For the CF, we solve d2 y + 2 8y = 0, and as before, we find the solution 2 y c = Ae 2x + Be 4x. For the PI, we see the forcing function 2e 3x on the right hand side of the ODE. So, we try a solution of the form = αe 3x, and determine α. To find α, we use the given ODE. Firstly, y p = 3αe 3x, and y p = 9αe 3x. Substituting these into the ODE, we get 9αe 3x + 2 3αe 3x 8αe 3x = 2e 3x 7α = 2 α = 2 7. So, = 2 7 e3x. The final, general solution, is y = Ae 2x + Be 4x e3x. In general, we try = αe kx and solve for α, if we have the forcing function ce kx. But, ce kx must not be part of the CF. In Example 7, this method would fail if 2e 3x is replaced by a function of the form ce 2x, or de 4x. We will come back to this problem in Examples 0 and. Example 8. Find the general solution of y = 3 sin 2x. 2 Solution. For the CF, same as before. We find that y c = (Ax + B)e 5x. For the PI, seeing the forcing function 3 sin 2x on the right, we try = α sin 2x + β cos 2x, and determine α and β. We do the same thing as before. = 2α cos 2x 2β sin 2x, = 4α sin 2x 4β cos 2x. Substituting these into the ODE, we get 4α sin 2x 4β cos 2x 0(2α cos 2x 2β sin 2x) + 25(α sin 2x + β cos 2x) = 3 sin 2x. 8

9 We balance out the sin and cos terms. Collecting up the sin terms, we find Collecting up the cos terms, we find 4α + 20β + 25α = 3, or, 2α + 20β = 3 (7) 4β + 20α + 25β = 0, or, 20α + 2β = 0 (8) Now, solve the simultaneous equations (7) and (8). We find that α = and β =. 4 4 For example, we can do this by subtracting (8) from (7), which gives α β = 3. Hence, 20α 20β = 60. Adding this to (7) gives 4α = 63, so α = 63. Then, β = α 3 = = So, = sin 2x cos 2x. The final, general solution, is 4 4 y = (Ax + B)e 5x sin 2x cos 2x. 4 4 In general, if we see the forcing function c sin kx + d cos kx, we try = α sin kx + β cos kx, and find α and β as above. But, there is an exception: If the auxiliary equation gives the complex roots ±iq, giving the CF A sin qx + B cos qx, and the forcing function contains sin qx or cos qx. We need to use something else for the PI in this case. We will see an example of this situation in Example 2. Example 9. Find the general solution of y = 34x2 + 2x 5. Solution. For the CF, again same as before. We find that y c = e 4x (A sin x + B cos x). For the PI, we see the forcing function 34x 2 + 2x 5 on the right, a polynomial of degree 2. We try = αx 2 + βx + γ, another polynomial of degree 2, and determine α, β and γ. We use a procedure similar to the last example. Substituting these into the ODE, we get = 2αx + β, and = 2α. 2α 8(2αx + β) + 7(αx 2 + βx + γ) = 34x 2 + 2x 5. We equate the coefficients of x 2, x, and the constant terms. Collecting up the x 2 terms, we find 7α = 34 α = 2. Collecting up the x terms, we find 6α + 7β = 2 7β = 2 + 6α = 34 β = 2. Equating the constants, we find 2α 8β + 7γ = 5 7γ = 5 2α + 8β = = 7 γ =

10 So, = 2x 2 + 2x + 7. The final, general solution, is 7 y = e 4x (A sin x + B cos x) + 2x 2 + 2x In general, if we see a polynomial a n x n + a n x n + + a x + a 0 for the forcing function, we try = α n x n + α n x n + + α x + α 0, and find α n, α n,..., α, α 0 with the above method. Example 0. Find the general solution of 2 d2 y 2 6y = 2e2x. Solution. As in Example 7, the CF is y c = Ae 2x + Be 3 2 x. For the PI, we cannot try = αe 2x, because this is in the form of the CF. Instead, we simply multiply by an extra x, and use = αxe 2x. We then just carry on as before. By the product rule, = αe 2x + 2αxe 2x, = 2αe 2x + 2αe 2x + 4αxe 2x = 4αe 2x + 4αxe 2x. Substituting these into the ODE, and then cancelling out e 2x from every term, we get 2(4α + 4αx) (α + 2αx) 6αx = 2 8α + 8αx α 2αx 6αx = 2 7α = 2 α = 2 7. So, = 2 7 xe2x. The final, general solution, is y = Ae 2x + Be 3 2 x xe2x. In general, we try = αxe kx and solve for α, if we see the forcing function ce kx, and ce kx clashes with the CF like in this example. Example. Find the general solution of y = 3e4x. Solution. As before, the CF is y c = (Ax + B)e 4x. For the PI, we cannot try = αe 4x, and not even = αxe 4x. Both of these are in the form of the CF. So, we simply try = αx 2 e 4x. As before, = 2αxe 4x + 4αx 2 e 4x, = 2αe 4x + 8αxe 4x + 8αxe 4x + 6αx 2 e 4x = 2αe 4x + 6αxe 4x + 6αx 2 e 4x. 0

11 Substituting these into the ODE and cancelling e 4x, we get 2α + 6αx + 6αx 2 8(2αx + 4αx 2 ) + 6αx 2 = 3 2α = 3 α = 3 2. So, = 3 2 x2 e 4x. The final, general solution, is y = (Ax + B)e 4x x2 e 4x. In general, we try = αx 2 e kx and solve for α, if we see the forcing function ce kx, and ce kx clashes with the CF like in this example. Example 2. Find the general solution of + 9y = 2 cos 3x. 2 Solution. As before, the CF is y c = A sin 3x + B cos 3x. For the PI, we cannot try = α sin 3x + β cos 3x, because this is in the form of the CF. Instead, we simply multiply by an extra x, and use = x(α sin 3x + β cos 3x). We then just carry on as before. By the product rule, = α sin 3x + β cos 3x + x(3α cos 3x 3β sin 3x), = 3α cos 3x 3β sin 3x + 3α cos 3x 3β sin 3x + x( 9α sin 3x 9β cos 3x) = 6α cos 3x 6β sin 3x 9x(α sin 3x + β cos 3x). Substituting these into the ODE, and collecting the sin terms, we get Collecting the cos terms, we get 6β 9xα + 9xα = 0 β = 0. 6α 9xβ + 9xβ = 2 β = 3. So, = x cos 3x. The final, general solution, is 3 y = A sin 3x + B cos 3x + x cos 3x. 3 In general, we try = x(α sin 3x + β cos 3x) and solve for α and β, if we see the forcing function c sin kx + d cos kx, and this clashes with the CF like in this example. Example 3. Find the general solution of 3 d2 y 2 + 4y = e2x (50x 3 + x). Solution. As before, the CF is y c = Ae 4x + Be 3 x. For the PI, we see the forcing function e 2x (50x 3 + x). This is an exponential

12 multiplied by a polynomial, and we can proceed as follows. We make the substitution y = e 2x z, and we will find an ODE relating z and x. By the product rule, we have = 2e2x z + e, = 2 4e2x 2x dz z + 2e 2x dz dz + 2e2x + d2 z e2x = 2 4e2x 2x dz z + 4e + d2 z e2x. 2 Substituting these into the ODE, we find that all the e 2x terms will cancel. We get ( 3 4z + 4 dz + d2 z ) ( + 2z + dz ) 2 4z = 50x 3 + x 3 d2 z dz z = 50x3 + x. (9) Now, we go on to find the PI of the ODE (9), just like in Example 9 (We do not need to worry about finding the CF of this ODE). So, try z p = αx 3 + βx 2 + γx + δ. Then z p = 3αx 2 + 2βx + γ, and z p = 6αx + 2β. Substituting into (9), and equating the coefficients of x 3, we get Equating the coefficients of x 2, we find 30α = 50 α = α + 30β = 0 30β = 69α = 69.5 = 345 β = = Equating the coefficients of x, we find ( 3.6α β + 30γ = 30γ = 8α 46β = Equating the constants, we find γ = = 5. ( 3.2β + 23γ + 30δ = 0 30δ = 6β 23γ = 6 δ = = ) = ) 23.5 = This gives = e 2x z p = e 2x (5x x2 + 5x 46 ). The final, general solution, is 5 ( y = Ae 4x + Be 3 x + e 2x 5x x2 + 5x 46 ). 5 In general, if we see a forcing function of the form e kx (polynomial in x), then we use this method of substituting y = e kx z to find the PI. 2

13 Example 4. Find the general solution of 3 + 2y = 2x + cosh x. 2 Solution. As before, the CF is y c = Ae x + Be 2x. With a combination of functions for the forcing function, we simply try a PI which is a combination. We have 2x + cosh x = 2x + 2 ex + 2 e x. For 2x, we try αx + β. For 2 ex, we try γxe x (since γe x clashes with the CF). For 2 e x we try δe x. So, for the PI, we try = αx + β + γxe x + δe x. We use the same procedure to find α, β, γ and δ, by comparing the like terms. We have = α + γe x + γxe x δe x, = γe x + γe x + γxe x + δe x = 2γe x + γxe x + δe x. We substitute these into the ODE and compare the like terms. Comparing the coefficients of x, we find that 2α = 2, so α =. Then, comparing the constant terms, we find that 3α + 2β = 0, so that 2β = 3α = 3, so β = 3 2. Now, comparing the ex terms, we find that 2γ + γx 3(γ + γx) + 2γx = 2 Comparing the e x terms, we find that γ = 2 γ = 2. δ 3( δ) + 2δ = 2 6δ = 2 δ = 2. So, = x xex + 2 e x. The final, general solution, is y = Ae x + Be 2x + x xex + 2 e x. Finally, as before, we can find the particular solution of a second order ODE, if we are given more information. We will need two pieces of information if we want to find a particular solution exactly, since there are two arbitrary constants in the general solution. Example 5. Find the particular solution of the ODE of Example 7: if y(0) = and y (0) = y = 2e3x, Solution. We have alrea seen that the general solution is y = Ae 2x + Be 4x e3x. Using y(0) =, we have = A + B + 2, so A + B = 5. Now, 7 7 y = 2Ae 2x 4Be 4x e3x (by differentiating the general solution), so y (0) = 2 gives 2 = 2A 4B + 6, so 7 2A 4B = 20, so A 2B = 0. So, solving 7 7 A + B = 5 7, A 2B = 0 7 3

14 simultaneously, we find that A = 0 and B = 5. Hence, the particular solution is 7 y = 5 7 e 4x e3x. 5. Second Order ODEs with Variable Coefficients We now look at some second ordered ODEs with variable coefficients. We will specifically consider those of the form ax 2 d2 y + bx + cy = f(x), (0) 2 where a, b and c are given, real constants, with a 0, and again f(x) is a given function of x, the forcing function of the ODE. Again, the final solution to (0) will consist of a complementary function (CF) and a particular integral (PI). The CF again comes from the solution to the case f(x) 0, namely, ax 2 d2 y + bx + cy = 0. () 2 To solve (), we let y c = x λ. This gives y c = λx λ and y c = λ(λ )x λ 2. Substituting into () and cancelling x λ then gives aλ(λ ) + bλ + c = 0, leading to the auxiliary equation aλ 2 + (b a)λ + c = 0. (2) Solving (2) again leads to three possible forms of the general solution of y c. If the solutions of (2) are two distinct real roots λ and λ 2, then the general solution to () is y = Ax λ + Bx λ 2. If the solution of (2) is one (repeated) real root λ, then the general solution to () is y = (A ln x + B)x λ. If the solutions of (2) are two complex conjugate roots p + iq and p iq, then the general solution to () is Example 6. The general solutions to y = x p (A sin(q ln x) + B cos(q ln x)). x 2 d2 y + 4x 4y = 0, 2 4x2 are, respectively d2 y 8x + 9y = 0, 2 x2 d2 y 2 7x + 20y = 0 y = Ax + Bx 4, y = (A ln x + B)x 3/2, y = x 4 (A sin(2 ln x) + B cos(2 ln x)). 4

15 For instance, in the first case, we solve the auxiliary equation λ 2 + 3λ 4 = 0 to get λ, λ 2 =, 4. Now, we go back to the ODE (0). Here, we shall only consider forcing functions of the form f(x) = cx k. Note that k may not necessarily be an integer. We would like to find the PI again, and our final solution will again be y = y c +. We first consider the case where cx k does not clash with the CF. Example 7. Find the general solution of 4x 2 d2 y 2 + 5x 3y = 2x3. Solution. As before, the CF is y c = Ax 3/4 + Bx. For the PI, with the forcing function 2x 3, which does not clash with the CF, we just try = αx 3. This gives = 3αx 2, and = 6αx. Substituting these into the ODE, we find that we can cancel x 3. This gives 4.6α + 5.3α 3α = 2 36α = 2 α = 8. So, = 8 x3. The final, general solution, is y = Ax 3/4 + Bx + 8 x3. In general, if the forcing function is of the form cx k and it does not clash with the CF, then we try = αx k and solve for α. Now, we consider the case where the forcing function f(x) = cx k clashes with the CF. Example 8. Find the general solution of 2x 2 d2 y 2 7x + 4y = 4x/2. Solution. As before, the CF is y c = Ax /2 + Bx 4. For the PI, we see that the forcing function 4x /2 does clash with the CF. In this case, we try = αx /2 ln x. This gives = 2 αx /2 ln x + αx /2. x = 2 αx /2 ln x + αx /2, = 4 αx 3/2 ln x + 2 αx /2. x 2 αx 3/2 = 4 αx 3/2 ln x. Substituting these into the ODE, we find that we can cancel x /2. This gives 2 ( ) ( ) 4 α ln x 7 2 α ln x + α + 4α ln x = 4 7α = 4 α =

16 So, = 4 7 x/2 ln x. The final, general solution, is y = Ax /2 + Bx x/2 ln x. In general, we try = αx k ln x and solve for α, if we see the forcing function cx k, and this clashes with the CF like in this example. There is one further possible clash. If the forcing function is cx k, and the CF is of the form (A ln x + B)x k. This can happen if the auxiliary equation has equal real roots. The method in Example 8 does not quite work. Example 9. Find the general solution of x 2 d2 y 2 + 5x + 4y = x 2. Solution. As before, the CF is y c = (A ln x + B)x 2. For the PI, we see the forcing function x 2. We cannot try = αx 2, and not even = αx 2 ln x, as both of these clash with the CF. We try = αx 2 (ln x) 2. This gives = 2αx 3 (ln x) 2 + αx 2. 2 ln x = 2αx 3 (ln x) 2 + 2αx 3 ln x, x = 6αx 4 (ln x) 2 2αx 3. 2 ln x 6αx 4 ln x + 2αx 3. x x = 6αx 4 (ln x) 2 0αx 4 ln x + 2αx 4. Substituting these into the ODE, we find that we can cancel x 2. This gives 6α(ln x) 2 0α ln x + 2α + 5( 2α(ln x) 2 + 2α ln x) + 4α(ln x) 2 = 2α = α = 2. So, = 2 x 2 (ln x) 2. The final, general solution, is y = (A ln x + B)x x 2 (ln x) 2. In general, we try = αx k (ln x) 2 and solve for α, if we see the forcing function cx k, and this clashes with the CF like in this example. 6