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)

y^{^{''}} + 2 y^{^{'}} + y = e^{3} + e^{2 x}

y^{^{''}} + y^{^{'}} + 12 y = 20 cos (3 t)

y^{^{''}} - 9 y = sin (x)

\frac{d ^{2} u}{d y ^{2}} - 4 u + 6 = 0

y^{^{''}} - 16 y = 32 + 45 e^{x}, y (0) = 0

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)