What is the solution for y^{''}+25y=sin(9x) ?

The solution for y^{''}+25y=sin(9x) is y=c_{1}cos(5x)+c_{2}sin(5x)-1/56 sin(9x)

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y^{''}+25y=sin(9x)

y^{^{''}} + 25 y = sin (9 x)

y = c_{1} cos (5 x) + c_{2} sin (5 x) - \frac{1}{56} sin (9 x)

Solution steps

y^{''} (x) + 25 y = sin (9 x)

Solve linear ODE:

y = c_{1} cos (5 x) + c_{2} sin (5 x) - \frac{1}{56} sin (9 x)

y = c_{1} cos (5 x) + c_{2} sin (5 x) - \frac{1}{56} sin (9 x)

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

y^{^{''}} + 9 y^{^{'}} = 18 e^{x} sin (x)

y^{^{''}} + 4 y = 3 cos (t)

y^{^{''}} + 3 y^{^{'}} + 1 = 0

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

What is the solution for y^{''}+25y=sin(9x) ?
The solution for y^{''}+25y=sin(9x) is y=c_{1}cos(5x)+c_{2}sin(5x)-1/56 sin(9x)