What is the solution for y^{''}+9y=4sin(3x) ?

The solution for y^{''}+9y=4sin(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)-(2xcos(3x))/3

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y^{''}+9y=4sin(3x)

y^{^{''}} + 9 y = 4 sin (3 x)

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

Solution steps

y^{''} (x) + 9 y = 4 sin (3 x)

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{2} \cdot e^{- 2 x} + 1

y^{^{''}} + 7 y = 5

x^{^{''}} + x^{^{'}} - 2 x = 8 sin (2 t)

y^{^{''}} - 5 y^{^{'}} + 6 y = (12 x - 7) e^{- x}

y^{^{''}} + 25 y = 20 sin (5 x) + x cos (x)

What is the solution for y^{''}+9y=4sin(3x) ?
The solution for y^{''}+9y=4sin(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)-(2xcos(3x))/3