What is the solution for y^{''}+25y=-40sin(5t) ?

The solution for y^{''}+25y=-40sin(5t) is y=c_{1}cos(5t)+c_{2}sin(5t)+4tcos(5t)

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y^{''}+25y=-40sin(5t)

y^{^{''}} + 25 y = - 40 sin (5 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t) + 4 t cos (5 t)

Solution steps

y^{''} (t) + 25 y = - 40 sin (5 t)

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) + 4 t cos (5 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t) + 4 t cos (5 t)

\frac{\partial}{\partial x} (4 arctan (x y))

\int 12 d x

x \to 0 lim (\frac{x}{tan ( 2 x )})

d er i v a t i v e x 16 - x^{2}

x \frac{d y}{d x} - y = x^{2} y^{2}

What is the solution for y^{''}+25y=-40sin(5t) ?
The solution for y^{''}+25y=-40sin(5t) is y=c_{1}cos(5t)+c_{2}sin(5t)+4tcos(5t)