What is the solution for x^{''}-5x^'+6x=(t^2+t)sin(t) ?

The solution for x^{''}-5x^'+6x=(t^2+t)sin(t) is

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x^{''}-5x^'+6x=(t^2+t)sin(t)

x^{^{''}} - 5 x^{^{'}} + 6 x = (t^{2} + t) sin (t)

x = c_{1} e^{3 t} + c_{2} e^{2 t} + \frac{t ^{2} sin ( t )}{10} + \frac{t ^{2} cos ( t )}{10} + \frac{9 t sin ( t )}{50} + \frac{3 t cos ( t )}{10} + \frac{9}{250} sin (t) + \frac{28}{125} cos (t)

Solution steps

x^{''} (t) - 5 x^{'} (t) + 6 x = (t^{2} + t) sin (t)

Solve linear ODE:

x = c_{1} e^{3 t} + c_{2} e^{2 t} + \frac{t ^{2} sin ( t )}{10} + \frac{t ^{2} cos ( t )}{10} + \frac{9 t sin ( t )}{50} + \frac{3 t cos ( t )}{10} + \frac{9}{250} sin (t) + \frac{28}{125} cos (t)

x = c_{1} e^{3 t} + c_{2} e^{2 t} + \frac{t ^{2} sin ( t )}{10} + \frac{t ^{2} cos ( t )}{10} + \frac{9 t sin ( t )}{50} + \frac{3 t cos ( t )}{10} + \frac{9}{250} sin (t) + \frac{28}{125} cos (t)

x \to - 1 - lim (g (x))

f (x) = \frac{3}{( 2 x + 1 ) ^{\frac{5}{2}}}

x y^{^{'}} - y = - x^{2} + 2 x + 2

\int \frac{x ^{2} + 1}{x ^{2} - x} d x

\frac{\partial}{\partial y} (8 x ln (x y))

What is the solution for x^{''}-5x^'+6x=(t^2+t)sin(t) ?
The solution for x^{''}-5x^'+6x=(t^2+t)sin(t) is