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

The solution for y^{''}+9y=sin(x)+sin(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)+1/8 sin(x)-(xcos(3x))/6

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

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

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{8} sin (x) - \frac{x cos ( 3 x )}{6}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{8} sin (x) - \frac{x cos ( 3 x )}{6}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{8} sin (x) - \frac{x cos ( 3 x )}{6}

y^{^{''}} + 4 y = t^{2} + 4 e^{t}, y (0) = 0, y^{^{'}} (0) = 9

y^{^{''}} + 4 y = t^{2} + 4 e^{t}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 3} e^{5 t}

y^{^{''}} + k^{2} y = k_{2} \cdot sin (k x)

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

What is the solution for y^{''}+9y=sin(x)+sin(3x) ?
The solution for y^{''}+9y=sin(x)+sin(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)+1/8 sin(x)-(xcos(3x))/6