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

Lösung

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)

Schritte zur Lösung

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

Solve linear ODE:

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)

\frac{d ^{2} x}{d t ^{2}} + x = 6 cos (2 t)

y^{^{''}} - 2 y^{^{'}} + y = (x + 1) e^{x}

2 y^{^{''}} - 2 y^{^{'}} - 4 y = 2 e^{2 x}

D (x) = - 6 x + 4200

y^{^{''}} + 4 y = - 2, y (\frac{π}{8}) = \frac{1}{2}, y^{^{'}} (\frac{π}{8}) = 2