hw7-solutions

c2 et 1 t 1 0 problem 4 we are again looking for a

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Unformatted text preview: u1 (t) = tet − et + t + c1 and u2 (t) = −e−t − te−t + c2 . Putting these facts together, we obtain x(t) = (t − 1 + te−t + c1 e−t ) = c1 e−t −1 1 + (−1 − t + c2 et ) 1 0 2t + te−t −1 1 . + + c2 et −1 − t 1 0 Problem 4: We are, again, looking for a solution of the form x(t) = u1 (t) x1 (t) + u2 (t) x2 (t) = u1 (t) e−t −1 0 . + u2 (t) e−t t 1 In order for the function x(t) to be a solution of the inhomogeneous ODE x′ (t) = −1 0 −1 x(t) + , −1 −1 t it is necessary and sufficient that u′1 (t) e−t −1 −1 0 . = + u′2 (t) e−t t t 1 This gives u′1 (t) = 0 and u′2 (t) = et . Therefore, u1 (t) = c1 3 and u2 (t) = et + c2 . Putting these facts together, we obtain x(t) = c1 e−t = c1 e−t 0 −1 + (1 + c2 e−t ) 1 t −1 −1 0 + . + c2 e−t t t 1 Problem 5: We are looking for a solution of the form x(t) = u1 (t) x1 (t) + u2 (t) x2 (t) = u1 (t) sin(t) cos(t) . + u2 (t) cos(t) − sin(t) In order for the function x(t) to be a solution of the inhomogeneous ODE x′ (t) = cos(t) 01 , x(t) + − sin(t) −1 0 it is necessary and sufficient that u′1 (t) cos(t) sin(t) cos(t) . = + u′2 (t) − sin(t) cos(t) − sin(t) This gives u′1 (t) = 1 and u′2 (t) = 0. Therefore, u1 (t) = t + c1 and u2 (t) = c2 . Putting these facts together, we obtain x(t) = (t + c1 ) = c1 cos(t) sin(t) + c2 − sin(t) cos(t) cos(t) sin(t) cos(t) + c2 +t . − sin(t) cos(t) − sin(t) 4...
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