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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = \frac{e ^{2 t}}{e ^{2 t} + 1}

y^{^{''}} - 3 y = e^{2 t} + t e^{2 t} + t^{2} e^{2 t}

y^{^{''}} + 4 y = t^{2} + 8 e^{t}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 (9 x - 2) e^{2 x}

y^{^{''}} + 25 y = 10 cos (5 x) + 15 sin (5 x)

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