What is the solution for x^{''}+4x=12cos(2t) ?

The solution for x^{''}+4x=12cos(2t) is x=c_{1}cos(2t)+c_{2}sin(2t)+3tsin(2t)

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x^{''}+4x=12cos(2t)

x^{^{''}} + 4 x = 12 cos (2 t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) + 3 t sin (2 t)

Solution steps

x^{''} (t) + 4 x = 12 cos (2 t)

Solve linear ODE:

x = c_{1} cos (2 t) + c_{2} sin (2 t) + 3 t sin (2 t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) + 3 t sin (2 t)

y^{^{''}} + 6 y^{^{'}} + 8 y = 16 t + 52

y^{^{''}} - y^{^{'}} - 6 y = 36 x, y (0) = 4, y^{^{'}} (0) = - 17

y^{^{''}} + 2 y = cos (t)

y^{^{''}} - y^{^{'}} = x^{2} + 2 x + 3

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 32 y = 6 + 3 x

What is the solution for x^{''}+4x=12cos(2t) ?
The solution for x^{''}+4x=12cos(2t) is x=c_{1}cos(2t)+c_{2}sin(2t)+3tsin(2t)