What is the solution for y^{''}+25y=-20sin(5t) ?

The solution for y^{''}+25y=-20sin(5t) is y=c_{1}cos(5t)+c_{2}sin(5t)+2tcos(5t)

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y^{''}+25y=-20sin(5t)

y^{^{''}} + 25 y = - 20 sin (5 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t) + 2 t cos (5 t)

Solution steps

y^{''} (t) + 25 y = - 20 sin (5 t)

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) + 2 t cos (5 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t) + 2 t cos (5 t)

\frac{d}{d x} (3 cos (3 x + 1))

\frac{\partial}{\partial u} (b ((\frac{( N u )}{V ( u )})^{(\frac{2}{3})}))

\int_{1}^{\infty} \frac{1}{x ^{2}} d x

ln (x + 1), 2

What is the solution for y^{''}+25y=-20sin(5t) ?
The solution for y^{''}+25y=-20sin(5t) is y=c_{1}cos(5t)+c_{2}sin(5t)+2tcos(5t)