What is the solution for y^{''}-4y=sin(3t) ?

The solution for y^{''}-4y=sin(3t) is y=c_{1}e^{2t}+c_{2}e^{-2t}-1/13 sin(3t)

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y^{''}-4y=sin(3t)

y^{^{''}} - 4 y = sin (3 t)

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{1}{13} sin (3 t)

Solution steps

y^{''} (t) - 4 y = sin (3 t)

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{1}{13} sin (3 t)

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{1}{13} sin (3 t)

y^{^{''}} - 9 y = 4 - 30 x^{2} + 9 x^{4}

\frac{d ^{2} y}{d x ^{2}} + y = sec^{2} (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = e^{- x} cos (2 x)

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 13 y = 5 cos (2 x)

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y = - 1

What is the solution for y^{''}-4y=sin(3t) ?
The solution for y^{''}-4y=sin(3t) is y=c_{1}e^{2t}+c_{2}e^{-2t}-1/13 sin(3t)