In each of the following pairs of problems, we first find a particular solution using the method of variation of parameters using the given fundamental set of solutions {x1, x2}, and then we find a solution of the specified initial value problem. For the method of variation of parameters we use Theorem 4.7.1 in the special case that P (t) = A is a constant matrix. If X(t) is a fundamental matrix for the homogeneous equation x0 = Ax, then Z −1 xp(t) = X(t) X (t)g(t) dt is a particular solution to the inhomogeneous equation x0 = Ax + g(t). −1 1 Note that X (t) = W (t) adj(X(t)), where adj(X(t)) is the adjugate matrix of X(t), and W (t) = det X(t) is the Wronskian. To solve the initial value problem once we have xp(t) we solve for c1 and c2 so that x(0) = c1x1(0) + c2x2(0) + xp(0) equals the prescribed initial value.

Problems 2 and 6 −3 1  1  1  1  2 x0 = x+ , x = e−2t , x = e−4t , x(0) = . 1 −3 4t 1 1 2 −1 −1 The Wronskian is W (t) = det X(t) = −2e−6t, so we have Z 1 Z −e−4t −e−4t  1  u(t) = X−1(t)g(t) dt = − e6t dt 2 −e−2t e−2t 4t 1 Z e2t(1 + 4t) 1 e2t(−1 + 4t) = dt = . 2 e4t(1 − 4t) 4 e4t( 1 − 2t) Thus, our particular solution will be 1 −1 + 4t + 1 − 2t 1  t  x (t) = X(t)u(t) = = . p 4 −1 + 4t − 1 + 2t 2 −1 + 3t

To find the solution of the initial value problem, we need to find c1 and c2 so that 1  1 1  0  2 c + c + = . 1 1 2 −1 2 −1 −1

We must have c1 = 3/4 and c2 = 5/4, so our solution is 3 5 1  3e−2t + 5e−4t + 2t  x(t) = x (t) + x (t) + x (t) = . 4 1 4 2 p 4 3e−2t − 5e−4t − 2 + 6t

1 Problems 3 and 7 −1 −2 e−t 1 −1 −1 x0 = x + , x = e−t , x = et , x(0) = . 0 1 t 1 0 2 1 1 The Wronskian is W (t) = det X(t) = 1, so we have

Z Z et et  e−t u(t) = X−1(t)g(t) dt = dt 0 e−t t Z 1 + tet  t + tet − et  = dt = . te−t −te−t − e−t

Thus, our particular solution will be

(t + tet − et)e−t + (−te−t − e−t)(−et) x (t) = X(t)u(t) = p (−te−t − e−t)et te−t + t − 1 + t + 1 te−t + 2t = = . −t − 1 −t − 1

To find the solution of the initial value problem, we need to find c1 and c2 so that 1 −1  0 −1 c + c + = . 1 0 2 1 −1 1

We must have c1 = 1 and c2 = 2, so our solution is

e−t − 2et + te−t + 2t x(t) = x (t) + 2x (t) + x (t) = . 1 2 p 2et − t − 1

Problems 4 and 8 −1 0 −1 0 −1 1 x0 = x + , x = e−t , x = e−t , x(0) = . −1 −1 t 1 1 2 t 0 The Wronskian is W (t) = det X(t) = e−2t, so we have

Z Z  t 1 −1 u(t) = X−1(t)g(t) dt = e2te−t dt −1 0 t Z  0   0  = dt = . et et

2 Thus, our particular solution will be

−1 x (t) = X(t)u(t) = . p t

To find the solution of the initial value problem, we need to find c1 and c2 so that 0 −1 −1 1 c + c + = . 1 1 2 0 0 0

We must have c1 = 0 and c2 = −2, so our solution is

 2e−t − 1  x(t) = −2x (t) + x (t) = . 2 p −2te−t + t

Problems 5 and 9  0 1  cos t  cos t sin t 1 x0 = x+ , x = , x = , x(0) = . −1 0 − sin t 1 − sin t 2 cos t 1 The Wronskian is W (t) = det X(t) = 1, so we have

Z Z cos t − sin t  cos t u(t) = X−1(t)g(t) dt = dt sin t cos t − sin t Z 1 t = dt = . 0 0

Thus, our particular solution will be

 t cos t x (t) = X(t)u(t) = . p −t sin t

To find the solution of the initial value problem, we need to find c1 and c2 so that 1 0 1 c + c = . 1 0 2 1 1

We must have c1 = c2 = 1, so our solution is

 cos t + sin t + t cos t x(t) = x (t) + x (t) + x (t) = . 1 2 p − sin t + cos t − t sin t

3