What is the solution for y^{''}+9y=sin(2x) ?

The solution for y^{''}+9y=sin(2x) is y=c_{1}cos(3x)+c_{2}sin(3x)+1/5 sin(2x)

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y^{''}+9y=sin(2x)

y^{^{''}} + 9 y = sin (2 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{5} sin (2 x)

Solution steps

y^{''} (x) + 9 y = sin (2 x)

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{5} sin (2 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{5} sin (2 x)

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} + y = 2 sin (3 x)

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

y^{^{'}} - y^{^{''}} = x + e^{- x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 5 e^{4 x}

y^{^{''}} + 4 y = x^{2} + 1

What is the solution for y^{''}+9y=sin(2x) ?
The solution for y^{''}+9y=sin(2x) is y=c_{1}cos(3x)+c_{2}sin(3x)+1/5 sin(2x)