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

The solution for y^{''}+4y^'=3sin(3t) is y=c_{1}+c_{2}e^{-4t}-3/25 sin(3t)-4/25 cos(3t)

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

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

y = c_{1} + c_{2} e^{- 4 t} - \frac{3}{25} sin (3 t) - \frac{4}{25} cos (3 t)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} - \frac{3}{25} sin (3 t) - \frac{4}{25} cos (3 t)

y = c_{1} + c_{2} e^{- 4 t} - \frac{3}{25} sin (3 t) - \frac{4}{25} cos (3 t)

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

y^{^{''}} - 3 y^{^{'}} + 5 = e^{2 x} - 3 cos (3 x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 3 x

x^{^{''}} + 9 x = sin (t), x (0) = 1, x^{^{'}} (0) = 1

(D^{2} + D) (y) = sin (x)

What is the solution for y^{''}+4y^'=3sin(3t) ?
The solution for y^{''}+4y^'=3sin(3t) is y=c_{1}+c_{2}e^{-4t}-3/25 sin(3t)-4/25 cos(3t)