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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

5 y^{^{''}} - 19 y^{^{'}} - 4 y = 1 + 4 x

(d^{3} - 7 d - 6) y = 2 e^{x} cos (\frac{x}{2})

\frac{1}{4} y^{^{''}} + 4 y = 4 tan (4 t) - \frac{1}{4} e^{3 t}

y^{^{''}} + 5 y^{^{'}} + 6 y = 3 e^{- 2 x}

y^{^{''}} - 4 y - 12 y = 12 t

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