What is the solution for x^{''}+25x=2cos(5t) ?

The solution for x^{''}+25x=2cos(5t) is x=c_{1}cos(5t)+c_{2}sin(5t)+(tsin(5t))/5

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x^{''}+25x=2cos(5t)

x^{^{''}} + 25 x = 2 cos (5 t)

x = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{t sin ( 5 t )}{5}

Solution steps

x^{''} (t) + 25 x = 2 cos (5 t)

Solve linear ODE:

x = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{t sin ( 5 t )}{5}

x = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{t sin ( 5 t )}{5}

x^{^{''}} + 9 x = sin (3 t)

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

y^{^{''}} - 4 y = 6 cos (3 t + 3 0^{\circ})

y^{^{''}} - 7 y^{^{'}} + 6 y = 3 e^{t}

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

What is the solution for x^{''}+25x=2cos(5t) ?
The solution for x^{''}+25x=2cos(5t) is x=c_{1}cos(5t)+c_{2}sin(5t)+(tsin(5t))/5