What is the solution for y^{''}+y^'=sin(3x) ?

The solution for y^{''}+y^'=sin(3x) is y=c_{1}+c_{2}e^{-x}-1/10 sin(3x)-1/30 cos(3x)

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y^{''}+y^'=sin(3x)

y^{^{''}} + y^{^{'}} = sin (3 x)

y = c_{1} + c_{2} e^{- x} - \frac{1}{10} sin (3 x) - \frac{1}{30} cos (3 x)

Solution steps

y^{''} (x) + y^{'} (x) = sin (3 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} - \frac{1}{10} sin (3 x) - \frac{1}{30} cos (3 x)

y = c_{1} + c_{2} e^{- x} - \frac{1}{10} sin (3 x) - \frac{1}{30} cos (3 x)

y^{^{''}} + y = 3 x^{2} + 4

y^{^{''}} + 4 y^{^{'}} + 5 y = 2 e^{- 4 x}, y (0) = 2, y^{^{'}} (0) = 3

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{6} cos (t)

(d^{2} + 3 d + 2) y = 1 + 3 x + x^{2}

x^{^{''}} + 3 x^{^{'}} - 4 x = - 4 t + 3

What is the solution for y^{''}+y^'=sin(3x) ?
The solution for y^{''}+y^'=sin(3x) is y=c_{1}+c_{2}e^{-x}-1/10 sin(3x)-1/30 cos(3x)