What is the solution for y^{''}+9y=sin(2x),y(0)=1,y^'(0)=0 ?

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

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y^{''}+9y=sin(2x),y(0)=1,y^'(0)=0

y^{^{''}} + 9 y = sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

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

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

\frac{d ^{2} y}{d x ^{2}} + 5 y = x sin (x)

y^{^{''}} + 25 y = 4 cos (5 x) - 7 e^{5 x}

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

y^{^{''}} + 2 y + y = e^{- t} ln (t)

y^{^{''}} - 10 y^{^{'}} + 25 y = t, y (0) = 0, y^{^{'}} (0) = 1

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