y^{''}-4y^'=(t^2+1)sin(2t)

解

y^{^{''}} - 4 y^{^{'}} = (t^{2} + 1) sin (2 t)

y = c_{1} e^{4 t} + c_{2} + \frac{t ^{2} cos ( 2 t )}{10} - \frac{t ^{2} sin ( 2 t )}{20} - \frac{7 t sin ( 2 t )}{50} - \frac{t cos ( 2 t )}{50} + \frac{4}{125} cos (2 t) - \frac{51}{1000} sin (2 t)

解答ステップ

y^{''} (t) - 4 y^{'} (t) = (t^{2} + 1) sin (2 t)

線形 ODE を解く:

y = c_{1} e^{4 t} + c_{2} + \frac{t ^{2} cos ( 2 t )}{10} - \frac{t ^{2} sin ( 2 t )}{20} - \frac{7 t sin ( 2 t )}{50} - \frac{t cos ( 2 t )}{50} + \frac{4}{125} cos (2 t) - \frac{51}{1000} sin (2 t)

y = c_{1} e^{4 t} + c_{2} + \frac{t ^{2} cos ( 2 t )}{10} - \frac{t ^{2} sin ( 2 t )}{20} - \frac{7 t sin ( 2 t )}{50} - \frac{t cos ( 2 t )}{50} + \frac{4}{125} cos (2 t) - \frac{51}{1000} sin (2 t)