A Second Course in Elementary Differential Equations: Problems and Solutions. Marcel B. Finan Arkansas Tech University c All Rights Reserved

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1 A Second Course in Elementary Differential Equations: Problems and Solutions Marcel B. Finan Arkansas Tech University c All Rights Reserved

2 Contents 8 Calculus of Matrix-Valued Functions of a Real Variable 4 9 nth Order Linear Differential Equations: Exsitence and Uniqueness 3 3 The General Solution of nth Order Linear Homogeneous Equations 5 3 Fundamental Sets and Linear Independence 3 Higher Order Homogeneous Linear Equations with Constant Coefficients 5 33 Non Homogeneous nth Order Linear Differential Equations 3 34 Existence and Uniqueness of Solution to Initial Value First Order Linear Systems Homogeneous First Order Linear Systems 5 36 First Order Linear Systems: Fundamental Sets and Linear Independence 6 37 Homogeneous Systems with Constant Coefficients Homogeneous Systems with Constant Coefficients: Complex Eigenvalues Homogeneous Systems with Constant Coefficients: Repeated Eigenvalues 84 4 NonHomogeneous First Order Linear Systems 96 4 Solving First Order Linear Systems with Diagonalizable Constant Coefficients Matrix 4 Solving First Order Linear Systems Using Exponential Matrix

3 43 The Laplace Transform: Basic Definitions and Results 7 44 Further Studies of Laplace Transform The Laplace Transform and the Method of Partial Fractions46 47 Laplace Transforms of Periodic Functions Convolution Integrals The Dirac Delta Function and Impulse Response Solving Systems of Differential Equations Using Laplace Transform 79 5 Numerical Methods for Solving First Order Linear Systems: Euler s Method 88 3

4 8 Calculus of Matrix-Valued Functions of a Real Variable Problem 8. Consider the following matrices [ t t A(t) t + (a) Find A(t) - 3tB(t) (b) Find A(t)B(t) - B(t)A(t) (c) Find A(t)c(t) (d) Find det(b(t)a(t)), B(t) [ t t + [ t +, c(t) (a) [ t t A(t) 3tB(t) t + (b) A(t)B(t) B(t)A(t) [ t t 4 4t + [ t 3t t + [ 3t 3t 3t + 6t [ t 3t t + 3t 4 t 3t [ t t t + [ t t + [ t t t 3 + t t + t t + 5t [ t t + [ t + t + 4 [ t t t + [ t t t 3 t t + 4 t + 5t + (c) [ t t A(t)c(t) t + [ t + [ (t )(t + ) + t ( ) (t + ) + (t + )( ) [ 4

5 (d) det(b(t)a(t)) [ [ [ t t t t t t 3 t t + t + t + 4 t + 5t + (t 3 + 3t + t) Problem 8. Determine all values t such that A(t) is invertible and, for those t-values, find A (t). [ t + t A(t) t t + We have det(a(t)) t + so that A is invertible for all t. In this case, A (t) [ t + t t + t t + Problem 8.3 Determine all values t such that A(t) is invertible and, for those t-values, find A (t). [ sin t cos t A(t) sin t cos t We have det(a(t)) sin t cos sin t so that A is invertible for all t nπ where n is an integer. In this case, A (t) sin t [ cos t cos t sin t sin t Problem 8.4 Find [ sin t 3 t cos t lim t t+ t e 3t t sec t t [ sin t 3 t cos t lim t t+ t e 3t t sec t t 5 [ 3

6 Problem 8.5 Find [ te t tan t lim t t e sin t [ te t tan t lim t t e sin t [ Problem 8.6 Find A (t) and A (t) if A(t) [ sin t 3t t + 5 Problem 8.7 Express the system in the matrix form [ cos t 3 A (t) t [ sin t A (t) y t y + 3y + sec t y (sin t)y + ty 5 y (t) A(t)y(t) + g(t) y(t) [ y (t) y (t) Problem 8.8 Determine A(t) where A (t) [ t 3, A(t) sin t t [ t cos t 3t [ sec t, g(t) 5 [ 5, A() 6

7 Integrating componentwise we find [ t A(t) + c t + c sin t + c t 3 + c Since A() [ 5 [ c c c c by equating componentwise we find c, c 5, c, and c. Hence, [ t A(t) + t + 5 sin t + t 3 + Problem 8.9 Determine A(t) where [ t A (t) [, A() [, A() 3 Integrating componentwise we find [ A t + t c (t) + c c c Integrating again we find A(t) [ t + c t + d t 3 + c 6 t + d c t + d c t + d But A() [ [ d d d d so by equating componentwise we find d, d, d, and d. Thus, [ t A(t) t + t3 + c 6 t + c t + c t + 7

8 Since A() [ 3 [ 3 + c 7 + c 6 + c + c we find c 5, c 5 6, c, c. Hence, A(t) [ t 5 t + t t + t + Problem 8. Calculate A(t) t B(s)ds where [ e B(s) s cos πs 6s sin πs Integrating componentwise we find [ t t A(t) es ds 6sds [ e t cos πsds t t 3t sin πt sin πsds π cos πt π Problem 8. Construct a nonconstant matrix function A(t) such that A (t) is a constant matrix. Let Then A (t) A(t) [ t [ t [ t [ Problem 8. (a) Construct a differentiable matrix function A(t) such that d dt A (t) A d dt A(t) That is, the power rule is not true for matrix functions. (b) What is the correct formula relating A (t) to A(t) and A (t)? 8

9 (a) Let Then so that A(t) A (t) [ t t [ + t 3 t t t 3 [ d 3t dt A (t) t 3t On the other hand, A(t) d [ [ t + 4t dt A(t) 3 t + t 4 t t + t 5 t 3 [ t t (b) (b) The correct formula is d dt A (t) A(t)A (t) + A (t)a(t) Problem 8.3 Transform the following third-order equation into a first order system of the form y 3ty + (sin t)y 7e t x (t) Ax(t) + b(t) Let x y, x y, x 3 y. Then by letting x x x, A x 3 sin t 3t, b 7e t then the differential equation can be presented by the first order system x (t) Ax(t) + b(t) Problem 8.4 By introducing new variables x and x, write y y + t as a system of two first order linear equations of the form x + Ax b 9

10 By letting x y and x y we have [ [ x x, A x [, b t Problem 8.5 Write the differential equation y + 4y + 4y as a first order system. By letting x y and x y we have [ [ x x, A 4 4 x [, b Problem 8.6 Write the differential equation y + ky + (t )y as a first order system. By letting x y and x y we have [ [ x x, A t k x [, b Problem 8.7 Change the following second-order equations to a first-order system. y 5y + ty 3t, y(), y () If we write the problem in the matrix form x + Ax b, x() y then x [ x x [ A 5 t [, b 3t [ y y [, y

11 Problem 8.8 Consider the following system of first-order linear equations. [ 3 x x Find the second-order linear differential equation that x satisfies. The system is It follows that x 3x + x x x x x + x 5x or x x +x 5 so we let w x +x so that w x 5. Thus, x 3x +x 3x +(5w x ) x + 5w. Hence, x x + 5w x + 5x or x x 5x Problem 8.9 List all the permutations of S {,, 3, 4} Problem 8. List all elementary products from the matrices (a) ( ) a a, a a (b) a a a 3 a a a 3 a 3 a 3 a 33

12 (a) The only elementary products are a a, a a. (b) An elementary product has the form a a a 3. Since no two factors come from the same column, the column numbers have no repetitions; consequently they must form a permutation of the set {,, 3}. The 3! 6 permutations yield the following elementary products: a a a 33, a a 3 a 3, a a 3 a 3, a a a 33, a 3 a a 3, a 3 a a 3 Problem 8. Find det(a) if (a) ( a a A a a ), (b) A a a a 3 a a a 3 a 3 a 3 a 33 By using the definition of a determinant and Exercise?? we obtain (a) A a a a a. (b) A a a a 33 a a 3 a 3 + a a 3 a 3 a a a 33 + a 3 a a 3 a 3 a a 3

13 9 nth Order Linear Differential Equations: Exsitence and Uniqueness For Problems , use Theorem 9. to find the largest interval a < t < b in which a unique solution is guaranteed to exist. Problem 9. y t 9 y + ln (t + ) + (cos t)y, y(), y () 3, y () The coefficient functions are all continuous for t 3,, 3. Since t, the largest interval of existence is < t < 3 Problem 9. y + t + y + (tan t)y, y(), y (), y () The coefficient functions are all continuous for t and t (n + ) π where n is an integer. Since t, the largest interval of existence is < t < π Problem 9.3 y t + 9 y + ln (t + )y + (cos t)y, y(), y () 3, y () The coefficient functions are all continuous for t so that the interval of existence is < t < Problem 9.4 Determine the value(s) of r so that y(t) e rt is a solution to the differential equation y y y + y 3

14 Inserting y and its derivatives into the equation we find y y y + y (r 3 r r + )e rt Since e rt >, we must have r 3 r r + r (r ) (r ) (r )(r ) (r )(r )(r + ). Hence, r,, Problem 9.5 Transform the following third-order equation into a first order system of the form y 3ty + (sin t)y 7e t x (t) Ax(t) + b(t) Let x y, x y, x 3 y. Then by letting x x x, A x 3 sin t 3t, b 7e t then the differential equation can be presented by the first order system x (t) Ax(t) + b(t) 4

15 3 The General Solution of nth Order Linear Homogeneous Equations In Problems , show that the given solutions form a fundamental set for the differential equation by computing the Wronskian. Problem 3. y y, y (t), y (t) e t, y 3 (t) e t We have W (t) e t e t e t e t e t e t () et e t e t e t et e t e t + e t e t e t Problem 3. y (4) + y, y (t), y (t) t, y 3 (t) cos t, y 4 (t) sin t We have W (t) t cos t sin t sin t cos t cos t sin t sin t cos t cos t + sin t Problem 3.3 t y + ty y, y (t), y (t) ln t, y 3 (t) t 5

16 We have W (t) () t t t ln t t t t t ln t t + t t t 3t, t > Use the fact that the solutions given in Problems for a fundamental set of solutions to solve the following initial value problems. Problem 3.4 y y, y() 3, y () 3, y () The general solution is y(t) c + c e t + c 3 e t. With the initial conditions we have y() 3 c + c + c 3 3 y () 3 c c 3 3 y () c + c 3 Solving these simultaneous equations gives c, c and c 3 and so the unique solution is Problem 3.5 y(t) e t + e t y (4) + y, y( π ) + π, y ( π ) 3, y ( π ) 3, y ( π ). The general solution is y(t) c + c t + c 3 cos t + c 4 sin t. With the initial conditions we have y( π ) + π c + π c + c 4 + π y ( π ) 3 c c 3 3 y ( π ) 3 c 4 3 y ( π ) c 3 6

17 Solving these simultaneous equations gives c (π + ), c 4, c 3 and c 4 3. Hence, the unique solution is Problem 3.6 y(t) (π + ) + 4t + cos t + 3 sin t t y + ty y,, y(), y (), y () 6 The general solution is y(t) c + c ln t + c 3 t. With the initial conditions we have y() c + c 3 y () c + c 3 y () 6 c + c 3 6 Solving these simultaneous equations gives c, c 4 and c 3 and so the unique solution is y(t) + 4 ln t t Problem 3.7 In each question below, show that the Wronskian determinant W (t) behaves as predicted by Abel s Theorem. That is, for the given value of t, show that W (t) W (t )e t t p n (s)ds (a) W (t) found in Problem 3. and t. (b) W (t) found in Problem 3. and t. (c) W (t) found in Problem 3.3 and t. (a) For the given differential equation p n (t) p (t) so that Abel s theorem predict W (t) W (t ). Now, for t we have W (t) W ( ) constant. From Problem 8., we found that W (t). (b) For the given differential equation p n (t) p 3 (t) so that Abel s theorem predict W (t) W (t ). Now, for t we have W (t) W () constant. 7

18 From Problem 8., we found that W (t). (c) For the given differential equation p n (t) p (t) so that Abel s t theorem predict W (t) W ()e t ds s W ()e ln ( t ) W (). From Problem t 8.3, we found that W (t) 3 so that W () 3 t Problem 3.8 Determine W (t) for the differential equation y +(sin t)y +(cos t)y +y such that W (). Here p n (t) p (t) sin t. By Abel s Theorem we have W (t) W ()e t sin sds Problem 3.9 Determine W (t) for the differential equation t 3 y y such that W () 3. Here p n (t) p (t). By Abel s Theorem we have Problem 3. Consider the initial value problem W (t) W ()e t ds W () 3 y y, y() α, y () β, y () 4. The general solution of the differential equation is y(t) c + c e t + c 3 e t. (a) For what values of α and β will lim t y(t)? (b) For what values α and β will the solution y(t) be bounded for t, i.e., y(t) M for all t and for some M >? Will any values of α and β produce a solution y(t) that is bounded for all real number t? (a) Since y() α, c + c + c 3 α. Since y (t) c e t c 3 e t and y () β, c c 3 β. Also, since y (t) c e t +c 3 e t and y () 4 we have c +c 3 4. Solving these equations for c, c, and c 3 we find c α 4, c β/ + and c 3 β/ +. Thus, y(t) α 4 + (β/ + )e t + ( β/ + )e t. 8

19 If α 4 and β 4 then y(t) 4e t and lim t 4e t. (b) In the expression of y(t) w know that e t is bounded for t whereas e t is unbounded for t. Thus, for y(t) to be bounded we must choose β/ + or β 4. The number α can be any number. Now, for the solution y(t) to be bounded on < t < we must have simultaneously β/ + and β/ +. But there is no β that satisfies these two equations at the same time. Hence, y(t) is always unbounded for any choice of α and β Problem 3. Consider the differential equation y + p (t)y + p (t)y on the interval < t <. Suppose it is known that the coefficient functions p (t) and p (t) are both continuous on < t <. Is it possible that y(t) c + c t + c 3 t 4 is the general solution for some functions p (t) and p (t) continuous on < t <? (a) Answer this question by considering only the Wronskian of the functions, t, t 4 on the given interval. (b) Explicitly determine functions p (t) and p (t) such that y(t) c +c t + c 3 t 4 is the general solution of the differential equation. Use this information, in turn, to provide an alternative answer to the question. (a) The Wronskian of, t, t 4 is W (t) t t 4 t 4t 3 t 6t3 Since is in the interval < t < and W (), {, t, t 4 } cannot be a fundamental set and therefore the general solution cannot be a linear combination of, t, t 4. (b) First notice that y is a solution for any p and p. If y t is a solution then substitution into the differential equation leads to tp + p. Since y t 4 is also a solution, substituting into the equation we find t p + 3tp + 6. Solving for p and p we find p (t) 3 t and p (t) 3 t. Note that both functions are not continuous at t 9

20 Problem 3. (a) Find the general solution to y. (b) Using the general solution in part (a), construct a fundamental set {y (t), y (t), y 3 (t)} satisfying the following conditions y (), y (), y (). y (), y (), y (). y (), y (), y (). (a) Using antidifferentiation we find that y(t) c + c t + c 3 t. (b) With the initial conditions of y we obtain the following system c + c + c 3 c + c 3 c 3 Solving this system we find c, c, c 3. Thus, y (t). Repeating this argument for y we find the system c + c + c 3 c + c 3 c 3 Solving this system we find c, c, c 3. Thus, y (t) t. Repeating this argument for y 3 we find the system c + c + c 3 c + c 3 c 3 Solving this system we find c, c, c 3. Thus, y 3(t) (t )

21 3 Fundamental Sets and Linear Independence Problem 3. Determine if the following functions are linearly independent First take derivatives The Wronskian is W (t) y (t) e t, y (t) sin (3t), y 3 (t) cos t y (t) e t y (t) 3 cos (3t) y 3(t) sin t y (t) 4e t y (t) 9 sin (3t) y 3(t) cos t e t sin 3t cos t e t 3 cos (3t) sin t 4e t 9 sin (3t) cos t e t ( 3 cos 3t cos t 9 sin 3t sin t) sin 3t( e t cos t + 4e t sin t) + cos t( 8e t sin 3t e t cos 3t) Thus,W () 5. Since this is a nonzero number, we can conclude that the three functions are linearly independent Problem 3. Determine whether the three functions : f(t), g(t) sin t, h(t) cos t, are linearly dependent or independent on < t < Computing the Wronskian W (t) sin t cos t sin t sin t cos t cos t [sin (t)( cos t) ( sin t( cos t)) So the functions are linearly depedent Problem 3.3 Determine whether the functions, y (t) ; y (t) + t; y 3 (t) + t + t ; are linearly dependent or independent. Show your work.

22 Equating coefficients we find ay + by + cy 3 a() + b( + t) + c( + t + t ) (a + b + c) + (b + c)t + ct a + b + c b + c c Solving this system we find that a b c so that y, y, and y 3 are linearly independent Problem 3.4 Consider the set of functions {y (t), y (t), y 3 (t)} {t +t, αt+, t+α}. For what value(s) α is the given set linearly depedent on the interval < t <? Equating coefficients we find ay + by + cy 3 a(t + t) + b(αt + ) + c(t + α) b + αc + (a + αb + c)t + at b + αc a + αb + c a Solving this system we find that a b c provided that α ±. In this case, y, y, and y 3 are linearly independent Problem 3.5 Determine whether the set {y (t), y (t), y 3 (t)} {t t +, t, t} is linearly independent or linearly dependent on the given interval (a) t <. (b) < t. (c) < t <.

23 (a) If t then y (t) t t + t +. In this case, we have Equating coefficients we find ay + by + cy 3 a(t + ) + b(t ) + c(t) (a + b)t + ct + a b a + b c a b Solving this system we find that a b c. This shows that y, y, and y 3 are linearly independent. (b) If t then y (t) t + (t ) y (t) + y 3 (t). Thus, y, y, and y 3 are linearly dependent. (c) Since y, y, and y 3 are linearly independent on the interval t <, they are linearly independent on the entire interval < t < In Problems , for each differential equation, the corresponding set of functions {y (t), y (t), y 3 (t)} is a fundamental set of solutions. (a) Determine whether the given set {y (t), y (t), y 3 (t)} is a solution set to the differential equation. (b) If {y (t), y (t), y 3 (t)} is a solution set then find the coefficient matrix A such that y y y 3 a a a 3 a a a 3 a 3 a 3 a 33 (c) If {y (t), y (t), y 3 (t)} is a solution set, determine whether it is a fundamental set by calculating the determinant of A. Problem 3.6 y y y 3 y + y {y (t), y (t), y 3 (t)} {, t, e t } {y (t), y (t), y 3 (t)} { t, t +, e (t+) } 3

24 (a) Since y (t) t, y (t) and y (t) y (t). Thus, y +y. Similarly, since y (t) t +, y (t) and y (t) y (t). Thus, y + y. Finally, since y 3 (t) e (t+), y 3(t) e (t+), y 3(t) e (t+) and y 3 (t) e (t+). Thus, y 3 + y 3. It follows, that {y (t), y (t), y 3 (t)} { t, t +, e (t+) } is a solution set. (b) Since y y y +y 3, y y +y +y 3, and y 3 y +y +e y 3 we have A e (c) Since det(a) 5e, {y (t), y (t), y 3 (t)} { t, t +, e (t+) } is a fundamental set of solutions Problem 3.7 t y + ty y, t > {y (t), y (t), y 3 (t)} {t, ln t, t } {y (t), y (t), y 3 (t)} {t, 3, ln (t 3 )} (a) Since y (t) t, y (t) 4t, y (t) 4, and y (t). Thus, t y + ty y. Similarly, since y (t) 3, y (t) y (t) y (t). Thus, t y + ty y. Finally, since y 3 (t) ln t 3, y 3(t) 3, t y 3(t) 3 and t y 3 (t) 6. Thus, t y t ty 3 y 3. It follows, that {y (t), y (t), y 3 (t)} {t, 3, ln t 3 } is a solution set. (b) Since y y +y +y 3, y 3y +y +y 3, and y 3 y +3y +y 3 we have 3 A 3 (c) Since det(a) 8, {y (t), y (t), y 3 (t)} {t, 3, ln t 3 } is a fundamental set of solutions 4

25 3 Higher Order Homogeneous Linear Equations with Constant Coefficients Problem 3. Solve y + y y y The characteristic equation is r 3 + r r. Factoring this equation using the method of grouping we find r (r + ) (r + ) (r + ) (r ) Hence, r is a root of multiplicity and r is a of multiplicity. Thus, the general solution is given by y(t) c e t + c te t + c 3 e t Problem 3. Find the general solution of 6y (4) 8y + y. The characteristic equation is 6r 4 8r +. This is a complete square Hence, r and r solution is given by (4r ) are roots of multiplicity. Thus, the general y(t) c e t + c te t + c3 e t + c4 te t Problem 3.3 Solve the following constant coefficient differential equation : y y. 5

26 In this case the characteristic equation is r 3 or r 3 e kπi. Thus, r e kπi 3 where k is an integer. Replacing k by,, and we find r r + i 3 r i 3 Thus, the general solution is y(t) c e t + c e 3t t cos + c 3e 3t t sin Problem 3.4 Solve y (4) 6y In this case the characteristic equation is r 4 6 or r 4 6 6e kπi. Thus, r e kπi where k is an integer. Replacing k by,, and 3 we find Thus, the general solution is r r i r r 3 i y(t) c e t + c e t + c 3 cos (t) + c 4 sin (t) Problem 3.5 Solve the initial-value problem y + 3y + 3y + y, y(), y (), y (). We have the characteristic equation r 3 + 3r + 3r + (r + ) 3 Which has a root of multiplicity 3 at r. We use what we have learned about repeated roots roots to get the general solution. Since the multiplicity of the repeated root is 3, we have y (t) e t, y (t) te t, y 3 (t) t e t. 6

27 The general solution is Now Find the first three derivatives y(t) c e t + c te t + c 3 t e t. y (t) c e t + c ( t)e t + c 3 (t t )e t y (t) c e t + c ( + t)e t + c 3 (t 4t + )e t Next plug in the initial conditions to get c c + c 3 Solving these equations we find c, c, and c 3. The unique solution is then y(t) te t + t e t Problem 3.6 Given that r is a solution of r 3 + 3r 4, find the general solution to y + 3y 4y Since r is a solution then using synthetic division of polynomials we can write (r )(r + ). Thus, the general solution is given by y(t) c e t + c e t + c 3 te t Problem 3.7 Given that y (t) e t is a solution to the homogeneous equation, find the general solution to the differential equation y y + y y The characteristic equation is given by r 3 r + r 7

28 Using the method of grouping we can factor into (r )(r + ) The roots are r, r i, and r 3 i. Thus, the general solution is y(t) c e t + c 3 cos t + c 4 sin t Problem 3.8 Suppose that y(t) c cos t + c sin t + c 3 cos (t) + c 4 sin (t) is the general solution to the equation Find the constants a, a, a, and a 3. y (4) + a 3 y + a y + a y + a y The characteristic equation is of order 4. The roots are given by r i, r i, r 3 i, and r 4 i. Hence the characteristic equation is (r + )(r + 4) or r 4 + 5r + 4. Comparing coefficients we find a 4, a, a 5, and a 3. Thus, the differential equation y (4) + 5y + 4y Problem 3.9 Suppose that y(t) c + c t + c 3 cos 3t + c 4 sin 3t is the general solution to the homogeneous equation y (4) + a 3 y + a y + a y + a y Determine the values of a, a, a, and a 3. The characteristic equation is of order 4. The roots are given by r (of multiplicity ), r 3i and r 3i. Hence the characteristic equation is r (r + 9) or r 4 + 9r. Comparing coefficients we find a a, a 9, and a 3. Thus, the differential equation y (4) + 9y Problem 3. Suppose that y(t) c e t sin t+c e t cos t+c 3 e t sin t+c 4 e t cos t is the general solution to the homogeneous equation y (4) + a 3 y + a y + a y + a y Determine the values of a, a, a, and a 3. 8

29 The characteristic equation is of order 4. The roots are given by r i, r +i, r 3 i, and r 4 +i. Hence (r ( i))(r ( +i)) r +r + and (r ( i))(r (+i)) r r + so that the characteristic equation is (r + r + )(r r + ) r Comparing coefficients we find a 4, a a a 3. Thus, the differential equation y (4) + 4y Problem 3. Consider the homogeneous equation with constant coefficients y (n) + a n y (n ) + + a y + a Suppose that y (t) t, y (t) e t, y 3 (t) cos t are several functions belonging to a fundamental set of solutions to this equation. What is the smallest value for n for which the given functions can belong to such a fundamental set? What is the fundamemtal set? The fundamental set must contain the functions y 4 (t) and y 5 (t) sin t. Thus, the smallest value of n is 5 and the fundamental set in this case is {, t, cos t, sin t, e t } Problem 3. Consider the homogeneous equation with constant coefficients y (n) + a n y (n ) + + a y + a Suppose that y (t) t sin t, y (t) e t sin t are several functions belonging to a fundamental set of solutions to this equation. What is the smallest value for n for which the given functions can belong to such a fundamental set? What is the fundamemtal set? The fundamental set must contain the functions y 3 (t) sin t, y 4 (t) cos t, y 5 (t) t sin t, y 6 (t) t cos t, y 7 (t) t cos t, and y 8 (t) e t cos t. Thus, the smallest value of n is 8 and the fundamental set in this case is {sin t, cos t, t sin t, t cos t, t sin t, t cos t, e t sin t, e t cos t} 9

30 Problem 3.3 Consider the homogeneous equation with constant coefficients y (n) + a n y (n ) + + a y + a Suppose that y (t) t, y (t) e t are several functions belonging to a fundamental set of solutions to this equation. What is the smallest value for n for which the given functions can belong to such a fundamental set? What is the fundamemtal set? The fundamental set must contain the functions y 3 (t) and y 4 (t) t Thus, the smallest value of n is 4 and the fundamental set in this case is {, t, t, e t } 3

31 33 Non Homogeneous nth Order Linear Differential Equations Problem 33. Consider the nonhomogeneous differential equation t 3 y + at y + bty + cy g(t), t > Determine a, b, c, and g(t) if the general solution is given by y(t) c t + c t + c 3 t 4 + ln t Since t, t, t 4 are solutions to the homogeneous equation then + + bt + ct b + c + at () + bt(t) + ct a + b + c t 3 (4t) + at (t ) + bt(4t 3 ) + ct 4 a + 4b + c 4 Solving the system of equations we find a 4, b 8, c 8. Thus, t 3 y 4t y + 8ty 8y g(t) But ln t is a particular solution so that g(t) t 3 (4t 3 4t ( t ) + 8t(t ) 6 ln t 8 6 ln t Problem 33. Consider the nonhomogeneous differential equation y + ay + by + cy g(t), t > Determine a, b, c, and g(t) if the general solution is given by y(t) c + c t + c 3 e t + 4 sin t The characteristic equation is r (r ) so that the associated homogeneous equation is y y. Thus, a, b c. The particular solution is y p (t) 4 sin t. Inserting into the equation we find g(t) 3 cos t ( 6 sin t) 3 cos t + 3 sin t 3

32 Problem 33.3 Solve y (4) + 4y 6 + 5e t We first find the solution to the homogeneous differential equation. characteristic equations is The roots are The homogeneous solution is r 4 + 4r or r (r + 4) r (repeated twice), r i, r i y h (t) c + c t + c 3 sin (t) + c 4 cos (t) The Since g(t) is a sum of two terms, we can work each term separately. The trial function for the function g(t) 6 is y p. Since this is a solution to the homogeneous equation, we multiply by t to get y p (t) t This is also a solution to the homogeneous equation, so multiply by t again to get y p (t) t which is not a solution of the homogeneous equation. We write y p At, y p At, y p A, y p A, y (4) p Substituting back in, we get + 4(A) 6 or A. Hence y p (t) t Now we work on the second piece. The trial function for g(t) 5e t is e t. Since this in not a solution to the homogeneous equation, we get y p Ae t, y p Ae t, y p Ae t, y p Ae t, y ( p 4) Ae t Plugging back into the original equation gives Ae t + 4Ae t 5e t and this implies A 3. Hence y p (t) 3e t The general solution to the nonhomogeneous differential equation is y(t) c + c t + c 3 sin (t) + c 4 cos (t) + t + 3e t 3

33 Problem 33.4 Solve: y (4) 8y + 6y 64e t The characteristic equation r 4 8r + 6 (r 4) has two double roots at r and r. The homogeneous solution is then y h (t) c e t + c te t + c 3 e t + c 4 te t Since the nonhomogeneous term is an exponential function, we use the method of undetermined coefficients to find a particular solution of the form Plug this into the equation and get y p (t) At e t (48A + 64At + 6At )e t 8(A + 8At + 4At )e t + 6At e t 64e t from which we see 3A 64 or A. The general solution is y(t) c e t + c te t + c 3 e t + c 4 te t t e t Problem 33.5 Given that y (t) e t is a solution to the homogeneous equation, find the general solution to the differential equation, y y + y y sin t The characteristic equation is r 3 r + r. We know that r is a root of this equation. Using synthetic division we can write r 3 r + r (r )(r + ) So the roots are r, r i, r i and the general solution is y h (t) c e t + c cos t + c 3 sin t To find a particular solution we use the method of undetermined coefficients by considering the trial function y p (t) A cos t + B sin t. In this case, we have y p(t) A sin t + B cos t y p(t) 4A cos t 4b sin t 33

34 Inseting into the differential equation we find sin t y p y p + y p y p 8A sin t 8B cos t ( 4A cos t 4B sin t) + (B cos t A sin t) (A cos t + B sin t) (6A 6B) cos t + (6A + 6B) sin t Equating coefficients we find 6A 6B and 6A + 6B. Solving we find A B and so the general solution is y(t) c e t + c cos t + c 3 sin t + cos t + sin t Problem 33.6 Find the general solution of the equation y 6y + y 8y te t The characteristic equation r 3 6r + r 8 (r ) 3 has a triple root at r. Hence, the homogeneous solution is y h (t) c e t + c te t + c 3 t e t We use the method of variation of paramemters to find the particular solution The Wronskian is W (t) y p (t) u e t + u te t + r 3 t e t e t te t t e t e t e t + te t te t + t e t 4e t 4e t + 4te t e t + 8te t + 4t e t e6t Also, W (t) W (t) te t t e t e t + te t te t + t e t 4e t + 4te t e t + 8te t + 4t e t e t t e t e t te t + t e t 4e t e t + 8te t + 4t e t t e 4t te4t 34

35 Hence, W 3 (t) e t te t e t e t + te t 4e t 4e t + 4te t e4t u (t) W (t) g(t)dt t 5 dt t 7 W (t) 7 u (t) W (t) g(t)dt t 3 dt W (t) u 3 (t) W3 (t) g(t)dt t dt t 3 W (t) 3 Hence, the general solution is 5 t 5 y(t) c e t + c te t + c 3 t e t + 7 t 7 e t 5 t 7 e t + 3 t 7 e t c e t + c te t + c 3 t e t t 7 e t Problem 33.7 (a) Verify that {t, t, t 4 } is a fundamental set of solutions of the differential equation t 3 y 4t y + 8ty 8y (b) Find the general solution of t 3 y 4t y + 8ty 8y t, t > (a) Let y (t) t, y (t) t, y 3 (t) t 4. Then t 3 y 4t y t 3 y 4t y t 3 y 3 4t y + 8ty 8y t 3 () 4t () + 8t() 8t + 8ty 8y t 3 () 4t () + 8t(t) 8t 3 + 8ty 3 8y 3 t 3 (4t) 4t (t ) + 8t(4t 3 ) 8t 4 t t t 4 W (t) t 4t 3 t 6t4, t > Since W () then {y, y, y 3 } is a fundamental set of solutions. (b) Using the method of variation of parameters we look for a solution of the form y p (t) u t + u t + u 3 t 4 35

36 where u (t) u (t) u 3(t) t t 4 t 4t 3 t t t 4 4t 3 t t 5 6t 4 3 t 3 t 5 t 5 6t 4 t t t t 5 6t 4 3 t 9 Thus, u (t) 3 t 3 dt 4 3 t u (t) t 5 dt 3 t 3 u 3 (t) 3 t 9 dt t 7 Thus, the general solution is y(t) c t + c t + c 3 t t + 3 t t c t + c t + c 3 t 4 6 t Problem 33.8 (a) Verify that {t, t, t 3 } is a fundamental set of solutions of the differential equation t 3 y 3t y + 6ty 6y (b) Find the general solution of by using the method of variation of parameters t 3 y 3t y + 6ty 6y t, t > (a) Let y (t) t, y (t) t, y 3 (t) t 3. Then t 3 y 3t y t 3 y 3t y t 3 y 3 3t y + 6ty 6y t 3 () 3t () + 6t() 6t + 6ty 6y t 3 () 3t () + 6t(t) 6t 3 + 6ty 3 6y 3 t 3 (6) 3t (6t) + 6t(3t ) 6t 3 t t t 3 W (t) t 3t 6t t3, t > 36

37 Since W () then {y, y, y 3 } is a fundamental set of solutions for t >. (b) Using the method of variation of parameters we look for a solution of the form y p (t) u t + u t + u 3 t 3 where u (t) u (t) u 3(t) t t 3 t 3t 6t t 3 t t 3 3t 6t t t t t 3 t t t t t t 3 t 3 Thus, u (t) dt ln t t u (t) dt t t u 3 (t) dt t 3 4t Thus, the general solution is y(t) c t + c t + c 3 t 3 + t ln t t c t + c t + c 3 t 3 + t ln t since 3 t is a solution to the homogeneous equation 4 Problem 33.9 Solve using the method of undetermined coefficients: y y 4 + cos t We first solve the homogeneous differential equation The characteristic equation is y y r 3 r 37

38 Factoring gives r(r )(r + ) Solving we find r, r and r. The homogeneous solution is y h (t) c + c e t + c 3 e t The trial function generated by g(t) 4 + cos (t) is Then y p (t) At + B cos (t) + C sin (t) y p A B sin (t) + C cos (t) y p 4B cos (t) 4C sin (t) y p 8B sin (t) 8C cos (t) Plugging back into the original differential equation gives [8B sin (t) 8C cos (t) [A B sin (t) + C cos (t) 4 + cos (t) Combining like terms gives Equating coefficients gives C cos (t) + B sin (t) A 4 + cos (t) C B A 4 Solving we find A 4, B, and C. The general solution is thus 5 y(t) c + c e t + c 3 e t 4t sin (t) 5 Problem 33. Solve using the method of undetermined coefficients: y y 4e t We first solve the homogeneous differential equation y y 38

39 The characteristic equation is Factoring gives r 3 r r(r )(r + ) Solving we find r, r and r. The homogeneous solution is y h (t) c + c e t + c 3 e t The trial function generated by g(t) 4e t is Then y p (t) Ate t y p Ae t + Ate t y p Ae t + Ate t y p 3Ae t + Ate t Plugging back into the original differential equation gives Combining like terms gives (3Ae t + Ate t ) (Ae t + Ate t ) 4e t Ae t 4e t Solving we find A. The general solution is thus y(t) c + c e t + c 3 e t te t Problem 33. Solve using the method of undetermined coefficients: y y 4e t The characteristic equation is r 3 r Factoring gives r (r ) 39

40 Solving we find r (double root) and r. The homogeneous solution is y h (t) c + c t + c 3 e t The trial function generated by g(t) 4e t is Then y p (t) Ae t y p Ae t y p 4Ae t y p 8Ae t Plugging back into the original differential equation gives Combining like terms gives ( 8e t ) (4Ae t ) 4e t Ae t 4e t Solving we find A. The general solution is thus 3 y(t) c + c e t + c 3 e t 3 e t Problem 33. Solve using the method of undetermined coefficients: y 3y +3y y e t. We first solve the homogeneous differential equation The characteristic equation is Factoring gives y 3y + 3y y r 3 3r + 3r (r ) 3 Solving we find r of multiplicity 3. The homogeneous solution is y h (t) c e t + c te t + c 3 t e t 4

41 The trial function generated by g(t) e t is Then y p (t) At 3 e t y p 3At e t + At 3 e t y p 6Ate t + 6At e t + At 3 e t y p 6Ae t + 8Ate t + 9At e t + At 3 e t Plugging back into the original differential equation gives (6Ae t +8Ate t +9At e t +At 3 e t ) 3(6Ate t +6At e t +At 3 e t )+3(3At e t +At 3 e t ) At 3 e t e t Combining like terms gives 6Ae t e t Solving we find A. The general solution is thus y(t) c e t + c te t + c 3 t e t + t 3 e t Problem 33.3 Solve using the method of undetermined coefficients: y + y e t + cos t. The characteristic equation is Factoring gives r 3 + (r + )(r r + ) Solving we find r, r i 3 and r + i 3. The homogeneous solution is y h (t) c e t + c e t 3 cos t + c 3e t 3 sin t The trial function generated by g(t) e t + cos t is y p (t) Ae t + B cos t + C sin t Then y p Ae t B sin t + C cos t y p Ae t B cos t C sin t y p Ae t + B sin t C cos t 4

42 Plugging back into the original differential equation gives or (Ae t + B sin t C cos t) + (Ae t + B cos t + C sin t) e t + cos t Ae t + (B + C) sin t + (B C) cos t e t + cos t Equating coefficients we find A B + C B C Solving we find A B and C. Thus, the general solution is y(t) c e t + c e t 3 cos t + c 3e t 3 sin t + (et + cos t sin t In Problems 33.4 and 33.5, answer the following two questions. (a) Find the homogeneous general solution. (b) Formulate an appropriate for for the particular solution suggested by the method of undetermined coefficients. You need not evaluate the undetermined coefficients. Problem 33.4 y 3y + 3y y e t + 4e t cos 3t + 4 (a) The characteristic equation is r 3 3r + 3r (r ) 3 So r is a root of multiplicity 3 so that the homogeneous general solution is y h (t) c e t + c te t + c 3 t e t. (b) The trial function for the right-hand function g(t) e t + 4e t cos 3t + 4 is y p (t) At 3 e t + Be t cos 3t + Ce t sin 3t + D. Problem 33.5 y (4) + 8y + 6y t cos t 4

43 (a) The characteristic equation is r 4 + 8r + 6 (r + 4) So r i and r i are roots of multiplicity so that the homogeneous general solution is y h (t) c cos t + c sin t + c 3 t cos t + c 4 t sin t. (b) The trial function for the right-hand function g(t) t cos t is y p (t) t (At + B) cos t + t (Ct + D) sin t. Consider the nonhomogeneous differential equation y + ay + by + cy g(t) In Problems , the general solution of the differential equation is given, where c, c, and c 3 represent arbitrary constants. Use this information to determine the constants a, b, c and the function g(t). Problem 33.6 y(t) c + c t + c 3 e t + 4 sin t. The roots of the characteristic equation are r of multiplicity and r. Thus, the equation is r (r ) r 3 r. The corresponding differential equation is y y. Comparing coefficients we find a, b c. The function y p (t) 4 sin t is a particular solution to the nonhomogeneous equation. Taking derivatives we find y p(t) 8 cos t, y p(t) 6 sin t, y p (t) 3 cos t. By plugging into the equation we find 3 cos t + 3 sin t g(t) Problem 33.7 y(t) c + c t + c 3 t t 3 43

44 The roots of the characteristic equation are r of multiplicity 3. Thus, the equation is r 3. The corresponding differential equation is y. Comparing coefficients we find a b c. The function y p (t) t 3 is a particular solution to the nonhomogeneous equation. Taking derivatives we find y p(t) 6t, y p(t) t, y p (t). By plugging into the equation we find g(t) Problem 33.8 Consider the nonhomogeneous differential equation t 3 y + at y + bty + cy g(t), t > Suppose that y(t) c t + c t + c 3 t 4 + ln t is the general solution to the above equation. Determine the constants a, b, c and the function g(t) The functions y (t) t, y (t) t, and y 3 (t) t 4 are solutions to the homogeneous equation. Substituting into the equation we find + + bt + ct b + c + at + bt + ct a + b + c 4t 4 + at 4 + 4bt 4 + ct 4 a + 4b + c 4 Solving this system of equations we find a 4, b 8, and c 8. Thus, t 3 y 4t y + 8ty 8y g(t). Since y p (t) ln t is a particular solution to the nonhomogeneous equation then by substitution we find g(t) t 3 ( 4 ) 4t ( ) + 8t( ) 6 ln t t 3 t t 8 6 ln t 44

45 34 Existence and Uniqueness of Solution to Initial Value First Order Linear Systems Problem 34. Consider the initial value problem (t + )y 3ty + 5y, y () (t )y y + 4ty, y () Determine the largest t-interval such that a unique solution is guaranteed to exist. All the coefficient functions and the right-hand side functions are continuous for all t ±. Since t, the t interval of existence is < t < Problem 34. Verify that the functions y (t) c e t cos t+c e t sin t and y (t) c e t sin t+ c e t cos t are solutions to the linear system y y + y y y + y Taking derivatives we find y (t) (c + c )e t cos t + (c c )e t sin t and y (t) (c +c )e t sin t+(c c )e t cos t. But y +y (c +c )e t cos t+(c c )e t sin t y (t) and y + y (c + c )e t sin t + (c c )e t cos t y (t) Problem 34.3 Consider the initial value problem where y (t) Ay(t), y() y [ 3 A 4 [, y 8 [ [ (a) Verify that y(t) c e 5t + c e t is a solution to the first order linear system. (b) Determine c and c such that y(t) solves the given initial value problem. 45

46 (a) We have y (t) c e 5t [ c e t [ and Ay(t) c e 5t [ 3 4 [ + c e t [ 3 4 [ (b) We need to solve the system [ 5 c e 5t 5 [ + c e t y (t) c c c + c 8 Solving this system we find c and c 3. Therefore, the unique solution to the system is [ e y(t) 5t 3e t e 5t + 6e t Problem 34.4 Rewrite the differential equation (cos t)y 3ty + ty t + in the matrix form y(t) P(t)y(t) + g(t). Rewriting the equation in the form y 3t sec ty + t sec t (t + ) sec t we find y [ y y [ t sec t 3t sec t [ y y P(t)y + g(t) [ + (t + ) sec t Problem 34.5 Rewrite the differential equation y + ty + e 3t y + (cos t)y in the matrix form y(t) P(t)y(t) + g(t). 46

47 Rewriting the given equation in the form y y + (cos t)y ty e 3t we find y y y y t cos t P(t)y(t) + g(t) Problem 34.6 The initial value problem [ y (t) 3 [ y + cos (t) y y y + [ e 3t [, y( ) 4 was obtained from an initial value problem for a higher order differential equation. What is the corresponding scalar initial value problem? Carrying the matrix arithmetic we find y 3y + y + cos t. Thus, the initial value problem is Problem 34.7 The initial value problem y + 3y y cos t, y( ), y ( ) 4 y (t) y y 3 y 4 y + y 3 sin y + y 3, y() was obtained from an initial value problem for a higher order differential equation. What is the corresponding scalar initial value problem? Let y(t) y y y 3 y 4 y y y y 47

48 Then y y y y y (4) Equating components we find Thus, the initial value problem is y y y y + y sin y + (y ) y (4) y + y sin y + (y ). y (4) y + y sin y + (y ), y() y (), y (), y () Problem 34.8 Consider the system of differential equations Write the above system in the form y tz + y + z z y + z + ty y P(t)y + g(t) where Identify P(t) and g(t). y(t) y(t) y (t) z(t) z (t) y y y z z t t y y z z P(t)y + G(t) 48

49 Problem 34.9 Consider the system of differential equations Write the above system in the form y 7y + 4y 8z + 6z + t z 5z + z 6y + 3y sin t y P(t)y + g(t) where Identify P(t) and g(t). y(t) y(t) y (t) z(t) z (t) y y y z z y y z z t sin t P(t)y + G(t) 49

50 35 Homogeneous First Order Linear Systems In Problems answer the following two questions. (a) Rewrite the given system of linear homogeneous differential equations as a homogeneous linear system of the form y (t) P(t)y. (b) Verify that the given function y(t) is a solution of y (t) P(t)y. Problem 35. and y 3y y y 4y + 3y [ y(t) e t + e t e t e t (a) (b) We have and P(t)y [ y [ y [ [ y [ e y t e t e t + e t [ e t + e t e t e t y [ e t e t e t + e t y Problem 35. and y y y y t + y t y(t) [ t + 3t t + 3 5

51 (a) (b) We have and Problem 35.3 P(t)y [ y y [ t t y [ t t [ t + 3 [ t + 3t t + 3 [ y y [ t + 3 y and y y + y + y 3 y y + y + y 3 y 3 y + y + y 3 y(t) e t + e 4t e t + e 4t e t + e 4t (a) (b) We have and P(t)y y y y 3 y e t + 4e 4t e t + 4e 4t e t + 4e 4t e t + e 4t e t + e 4t e t + e 4t y y y 3 e t + 4e 4t e t + 4e 4t e t + 4e 4t y In Problems (a) Verify the given functions are solutions of the homogeneous linear system. 5

52 (b) Compute the Wronskian of the solution set. On the basis of this calculation can you assert that the set of solutions forms a fundamental set? (c) If the given solutions are shown in part(b) to form a fundamental set, state the general solution of the linear homogeneous system. Express the general solution as the product y(t) Ψ(t)c, where Ψ(t) is a square matrix whose columns are the solutions forming the fundamental set and c is a column vector of arbitrary constants. (d) If the solutions are shown in part (b) to form a fundamental set, impose the given initial condition and find the unique solution of the initial value problem. Problem 35.4 [ [ 9 4 y y, y() 5 7 (a) We have and [ Similarly, and [ [ e, y (t) 3t 4e t 3e 3t e t [ 6e y 3t + 4e t 9e 3t + e t [ [ e 3t 4e t 6e 3e 3t e t 3t + 4e t 9e 3t + e t y [ e y 3t e t 8e 3t 5e t (b) The Wronskian is given by W (t) e3t 4e t 3e 3t e t [ [ 4e 3t + e t e 6e 3t + 5e t 3t e t 8e 3t 5e t y 4e 3t + e t 6e 3t + 5e t et [ 4e, y (t) 3t + e t 6e 3t + 5e t Since W (t), the set {y, y } forms a fundamental set of solutions. (c) The general solution is [ [ e y(t) c y + c y 3t 4e t 4e 3t + e t c 3e 3t e t 6e 3t + 5e t 5 c

53 (d) We have [ 6 7 [ c c Solving this system we find c.3, c.. Therefore the solution to the initial value problem is [ [ [ e y(t).3 3t 4e t 4e 3e 3t e t. 3t + e t e 6e 3t + 5e t 3t + e t.5e 3t +.5e t Problem 35.5 [ 3 5 y (a) We have and [ 3 5 Similarly, and [ 3 5 [ [ [ 5 y, y() y 5e t cos 3t e t (cos 3t 3 sin 3t) y 5e t sin 3t e t (3 cos 3t + sin 3t) [ [, y (t) [ y (t) [ 5e t ( cos 3t + 3 sin 3t) e t (cos 3t + sin 3t) [ 5e t ( sin 3t 3 cos 3t) e t ( 3 cos 3t sin 3t) 5e t cos 3t e t (cos 3t 3 sin 3t) 5e t sin 3t e t (3 cos 3t + sin 3t) [ 5e t ( cos 3t + 3 sin 3t) e t (cos 3t + sin 3t) [ 5e t ( sin 3t 3 cos 3t) e t ( 3 cos 3t sin 3t), y y (b) The Wronskian is given by W (t) 5e t cos 3t 5e t sin 3t e t (cos 3t 3 sin 3t) e t (3 cos 3t sin 3t) 5e 4t 53

54 Since W (t), the set {y, y } forms a fundamental set of solutions. (c) The general solution is [ [ 5e y(t) c y + c y t cos 3t 5e t sin 3t c e t (cos 3t 3 sin 3t) e t (3 cos 3t sin 3t) c (d) We have [ 5 3 [ c c Solving this system we find c, c. Therefore the solution to the initial value problem is [ [ [ 5e y(t) t cos 3t 5e e t t sin 3t 5e (cos 3t 3 sin 3t) e t t (cos 3t sin 3t) (3 cos 3t + sin 3t) e t ( cos 3t + 4 sin 3t) [ 5 Problem 35.6 [ y [ y, y( ) 4 [, y (t) [ e 3t, y (t) e 3t (a) We have and [ Similarly, and [ y [ [ y [ e 3t e 3t [ 3e 3t 6e 3t [ [ y 3e 3t 6e 3t (b) The Wronskian is given by W (t) e3t e 3t 3e 3t y 54

55 Since W (t), the set {y, y } forms a fundamental set of solutions. (c) The general solution is [ [ e 3t c y(t) c y + c y e 3t (d) We have [ e 3 e 3 [ c c [ 4 Solving this system we find c, c e 3. Therefore the solution to the initial value problem is [ e 3(t+) y(t) 4e 3(t+) c Problem 35.7 y 4 y, y() 3 4, y (t) e t, y (t) y 3 (t) e t cos t e t sin t e t sin t e t cos t (a) We have and Similarly, 4 y y e t e t e t e t (cos t sin t) e t (sin t + cos t) y 55

56 and and 4 4 e t cos t e t sin t y 3 e t cos t e t sin t e t (sin t + cos t) e t (cos t sin t) e t (cos t sin t) e t (sin t + cos t) e t (sin t + cos t) e t (cos t sin t) y y 3 (b) The Wronskian is given by e t W (t) e t cos t e t sin t e t sin t e t cos t. Since W (t), the set {y, y, y 3 } forms a fundamental set of solutions. (c) The general solution is e t c y(t) c y + c y + c 3 y 3 e t cos t e t sin t c e t sin t e t cos t c 3 (d) We have c c c 3 Solving this system using Cramer s rule we find c 3, c, c 3. Therefore the solution to the initial value problem is 3e t 3e t y(t) + 4e t cos t e t sin t 4e t sin t e t cos t 4e t (cos t sin t) e t (sin t + cos t) In Problems , the given functions are solutions of the homogeneous linear system

57 (a) Compute the Wronskian of the solution set and verify the set is a fundamental set of solutions. (b) Compute the trace of the coefficient matrix. (c) Verify Abel s theorem by showing that, for the given point t, W (t) W (t )e t t tr(p(s))ds. Problem 35.8 [ 6 5 y 7 6 [ 5e t y, y (t) 7e t [ e t, y (t) e t, t, < t < (a) The Wronskian is W (t) 5e t 7e t e t e t. Since W (t), the set {y, y } forms a fundamental set of solutions. (b) tr(p(t)) 6 6. t t (c) W (t) and W (t )e tr(p(s))ds e t ds Problem 35.9 [ t y t [ y, y (t) t [ e t, y (t), t, t, < t < (a) The Wronskian is W (t) et t t e t. Since W (t), the set {y, y } forms a fundamental set of solutions. (b) tr(p(t)) t t (c) W (t) t e t t and W (t )e tr(p(s))ds e e t ( s )ds t e t Problem 35. The functions y (t) [ 5 [ e 3t, y (t) e 3t 57

58 are known to be solutions of the homogeneous linear system y Py, where P is a real constant matrix. (a) Verify the two solutions form a fundamental set of solutions. (b) What is tr(p)? (c) Show that Ψ(t) satisfies the homogeneous differential equation Ψ PΨ, where [ 5 e 3t Ψ(t) [y (t) y (t) e 3t (d) Use the observation of part (c) to determine the matrix P.[Hint: Compute the matrix product Ψ (t)ψ (t). It follows from part (a) that Ψ (t) exists. Are the results of parts (b) and (d) consistent? (a) The Wronskian is given by W (t) 5 e 3t Since e3t 3e3t W (t), the set {y, y } forms a fundamental set of solutions. (b) Since W (t) tr(p(t))w (t), 9e 3t 3tr(P(t))e 3t. Thus, tr(p(t)) 3. (c) We have Ψ (t) [y y [P(t)y P(t)y P(t)[y y P(t)Ψ(t) (d) From part(c) we have [ 6e P(t) Ψ (t)ψ 3t (t) 3e 3t [ 5 [ e 3t e 3t 3e 3t 5 The results in parts (b) and (d) are consistent since tr(p(t)) Problem 35. The homogeneous linear system [ 3 y α y 58

59 has a fundamental set of solutions whose Wronskian is constant, W (t) 4, < t <. What is the value of α? We know that W (t) satisfies the equation W (t) tr(p(t))w (t). But tr(p(t)) 3 + α. Thus, W (t) (3 + α)w (t). Solving this equation we find W (t) W ()e t (3+α)ds. Since W (t) W () 4, e t (3+α)ds. Evaluating the integral we find e (3+α)t. This implies that 3 + α or α 3 59

60 36 First Order Linear Systems: Fundamental Sets and Linear Independence In Problems , determine whehter the given functions are linearly dependent or linearly independent on the interval < t <. Problem 36. f (t) [ t [ t, f (t) Suppose k [ t + k [ t [ Then k t+k t and k +k for all t. In particular, for t we have k + k. But k + k. These two equations imply that k k. Hence, {f (t), f (t)} is a linearly independent set Problem 36. [ e t f (t) [ e t, f (t) [ e t e t, f 3 (t) Note that [ e t [ e t [ e t e t This shows that f (t), f (t), and f 3 (t) are linearly dependent Problem 36.3 f (t) t, f (t) t, f 3 (t) 6

61 Note that t + t + This shows that f (t), f (t), and f 3 (t) are linearly dependent Problem 36.4 f (t) sin t, f (t) ( cos t), f 3 (t) Note that sin t ( cos t) This shows that f (t), f (t), and f 3 (t) are linearly dependent Problem 36.5 Consider the functions [ t f (t) [ t, f (t) (a) Let Ψ(t) [f (t) f (t). Determine det(ψ(t)). (b) Is it possible that the given functions form a fundamental set of solutions for a linear system y P(t)y where P(t) is continuous on a t-interval containing the point t? Explain. (c) Determine a matrix P(t) such that the given vector functions form a fundamental set of solutions for y P(t)y. On what t-interval(s) is the coefficient matrix P(t) continuous?(hint: The matrix Ψ(t) must satisfy Ψ (t) P(t)Ψ(t) and det(ψ(t)).) (a) We have F (t) t t 6 t.

62 (b) Since F (), the given functions do not form a fundamental set for a linear system y P(t)y on any t interval containing. (c) For Ψ(t) to be a fundamental matrix it must satisfy the differential equation Ψ (t) P(t)Ψ(t) and the condition det(ψ(t)). But det(ψ(t)) t and this is not zero on any interval not containing zero. Thus, our coefficient matrix P(t) must be continuous on either < t < or < t <. Now, from the equation Ψ (t) P(t)Ψ(t) we can find P(t) Ψ (t)ψ (t). That is, P(t) Ψ (t)ψ (t) t [ t [ t [ t t Problem 36.6 Let y y, Ψ(t) e t e t 4e t e t e t 3e t, Ψ(t) e t + e t 4e t e t + 4e t e t e t e t 3e t 3e t (a) Verify that the matrix Ψ(t) is a fundamental matrix of the given linear system. (b) Determine a constant matrix A such that the given matrix Ψ(t) can be represented as Ψ(t) Ψ(t)A. (c) Use your knowledge of the matrix A and assertion (b) of Theorem 36.4 to determine whether Ψ(t) is also a fundamental matrix, or simply a solution matrix. (a) Since and P(t)Ψ(t) Ψ (t) e t e t 8e t e t e t 6e t e t e t 4e t e t e t 3e t e t e t 8e t e t e t 6e t 6

63 Thus, Ψ is a solution matrix. To show that Ψ(t) is a fundamental matrix we need to verify that det(ψ(t)). Since det(ψ(t)) 6e t, Ψ(t) is a fundamental matrix. (b) Note that Ψ(t) Thus, e t + e t 4e t e t + 4e t e t e t e t 3e t 3e t A (c) Since det(a), Ψ(t) is a fundamental matrix Problem 36.7 Let y [ e t e t 4e t e t e t 3e t [ e t e y, Ψ(t) t 3e t where the matrix Ψ(t) is a fundamental matrix of the given homogeneous linear system. [ Find a constant matrix A such that Ψ(t) Ψ(t)A with Ψ(). We need to find [ a matrix A such that Ψ() Ψ()A or A Ψ ()Ψ() 3 Ψ () 3 63

64 37 Homogeneous Systems with Constant Coefficients In Problems , a matrix P and vectors x and x are given. (a) Decide which, if any, of the given vectors is an eigenvector of P, and determine the corresponding eigenvalue. (b) For the eigenpair found in part (a), form a solution y k (t), where k or k, of the first order system y Py. (c) If two solution are found in part (b), do they form a fundamental set of solutions for y Py. Problem 37. P [ [ 3, x 8 [, x (a) We have Px [ [ 3 8 x. Thus, x is an eigenvector corresponding to the eigenvalue r. Similarly, [ [ 7 3 Px x 6 7. Thus, x is an eigenvector corresponding to the eigenvalue r. (b) Solutions to the system y P(t)y are y (t) e t x and y (t) e t x. (c) The Wronskian is W (t) 3e t e t 8e t e t so that the set {y, y } forms a fundamental set of solutions Problem 37. P [ [, x 3 [, x 64

65 (a) We have Px [ [ 3 x. Thus, x is an eigenvector corresponding to the eigenvalue r. Similarly, [ [ [ 5 Px Since the right-hand side cannot be a scalar multiple of x, x is not an eigenvector of P. (b) Solution to the system y P(t)y is y (t) e t x. (c) The Wronskian is not defined for this problem Problem 37.3 P [ 4 [, x [, x (a) We have Px [ 4 [ 4x. Thus, x is an eigenvector corresponding to the eigenvalue r 4. Similarly, [ [ Px x 4. Thus, x is an eigenvector corresponding to the eigenvalue r. (b) Solutions to the system y P(t)y are y (t) e 4t x and y (t) x. (c) The Wronskian is W (t) e 4t e 4t 4e4t so that the set {y, y } forms a fundamental set of solutions In Problems , an eigenvalue is given of the matrix P. Determine a corresponding eigenvector. 65