What is the solution for y^{''}+9y=sin(2x),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+9y=sin(2x),y(0)=0,y^'(0)=0 is y=-2/15 sin(3x)+1/5 sin(2x)

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y^{''}+9y=sin(2x),y(0)=0,y^'(0)=0

y^{^{''}} + 9 y = sin (2 x), y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{2}{15} sin (3 x) + \frac{1}{5} sin (2 x)

Solution steps

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

Solve linear ODE:

y = - \frac{2}{15} sin (3 x) + \frac{1}{5} sin (2 x)

y = - \frac{2}{15} sin (3 x) + \frac{1}{5} sin (2 x)

y^{^{''}} + 4 y = 8 sin (2 x) + 5 e^{x}

y^{^{''}} - y^{^{'}} - 6 y = - e^{x} + 12 x, y (0) = 1, y^{^{'}} (0) = - 2

y^{^{''}} + 25 y = 82 e^{4 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = - 3 e^{2 x} + \frac{e ^{2 x}}{x}

y^{^{''}} + 8 y^{^{'}} + 12 y = 1 - H (t - 3)

What is the solution for y^{''}+9y=sin(2x),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+9y=sin(2x),y(0)=0,y^'(0)=0 is y=-2/15 sin(3x)+1/5 sin(2x)