What is the solution for y^{''}-4y^'=(t^2+1)sin(2t) ?

The solution for y^{''}-4y^'=(t^2+1)sin(2t) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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)

Solution steps

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)

What is the solution for y^{''}-4y^'=(t^2+1)sin(2t) ?
The solution for y^{''}-4y^'=(t^2+1)sin(2t) is