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

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

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y^{''}+9y=5sin(2x),y(0)=1,y^'(0)=0

y^{^{''}} + 9 y = 5 sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (x) + 9 y = 5 sin (2 x)

Solve linear ODE:

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

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

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