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

The solution for y^{''}+y=-sin(3x) is y=c_{1}cos(x)+c_{2}sin(x)+1/8 sin(3x)

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

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

y = c_{1} cos (x) + c_{2} sin (x) + \frac{1}{8} sin (3 x)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + \frac{1}{8} sin (3 x)

y = c_{1} cos (x) + c_{2} sin (x) + \frac{1}{8} sin (3 x)

\frac{d}{d x} (2 sin (π x))

\frac{d}{d x} (e^{\frac{4 x}{3}})

(x^{- 2} + x^{- 3}) \cdot (x^{5} - 2 x^{2})

\frac{d y}{d t} = y (3 - y)

\frac{d}{d x} (\frac{2 x}{x + 1})

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