What is the solution for (d^2+4d)y=2sin(3x) ?

The solution for (d^2+4d)y=2sin(3x) is y=c_{1}+c_{2}e^{-4x}-2/25 sin(3x)-8/75 cos(3x)

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(d^2+4d)y=2sin(3x)

(d^{2} + 4 d) y = 2 sin (3 x)

y = c_{1} + c_{2} e^{- 4 x} - \frac{2}{25} sin (3 x) - \frac{8}{75} cos (3 x)

Solution steps

(d^{2} + 4 d) y = 2 sin (3 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} - \frac{2}{25} sin (3 x) - \frac{8}{75} cos (3 x)

y = c_{1} + c_{2} e^{- 4 x} - \frac{2}{25} sin (3 x) - \frac{8}{75} cos (3 x)

y^{^{''}} + 9 y = 27 x^{2}

(d^{4} - 2 d^{3} + d^{2}) y = x^{3}

d^{2} (d^{2} + 4) y = 96 x^{2} + sin^{2} (x) - k

(y^{^{''}} - 4 y^{^{'}} + 3 y) = 2 x + 5

(d^{2} + 3 d + 2) y = (e^{e})^{x}

What is the solution for (d^2+4d)y=2sin(3x) ?
The solution for (d^2+4d)y=2sin(3x) is y=c_{1}+c_{2}e^{-4x}-2/25 sin(3x)-8/75 cos(3x)