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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{4 x}{( x ^{2} + 1 )}

\int_{0}^{3} (1 + 6 x^{2} - 10 x^{4}) d x

(1 + p x)^{- 1}

\frac{8}{ln ( x )}

sec^{2} (x)

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