What is the solution for (D^2+9)y=4cos(3x) ?

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

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(D^2+9)y=4cos(3x)

(D^{2} + 9) y = 4 cos (3 x)

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

Solution steps

(D^{2} + 9) y = 4 cos (3 x)

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} = 2 x + 9 - e - 2 x

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 t^{2} + 4 t e^{2 t} + t sin (2 t)

y^{^{''}} - 12 y^{^{'}} + 36 y = (x + 1) e^{x}

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- t}}{t + 1}

y^{^{'''}} - y^{^{'}} = (x + e^{x})^{2}

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