What is the solution for 4x^{''}+x=t^2+2cos(3t) ?

The solution for 4x^{''}+x=t^2+2cos(3t) is x=c_{1}cos(t/2)+c_{2}sin(t/2)+t^2-8-2/35 cos(3t)

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4x^{''}+x=t^2+2cos(3t)

4 x^{^{''}} + x = t^{2} + 2 cos (3 t)

x = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + t^{2} - 8 - \frac{2}{35} cos (3 t)

Solution steps

4 x^{''} (t) + x = t^{2} + 2 cos (3 t)

Solve linear ODE:

x = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + t^{2} - 8 - \frac{2}{35} cos (3 t)

x = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + t^{2} - 8 - \frac{2}{35} cos (3 t)

What is the solution for 4x^{''}+x=t^2+2cos(3t) ?
The solution for 4x^{''}+x=t^2+2cos(3t) is x=c_{1}cos(t/2)+c_{2}sin(t/2)+t^2-8-2/35 cos(3t)