What is the solution for y^{''}+25y=40cos(5t),y(0)=y ?

The solution for y^{''}+25y=40cos(5t),y(0)=y is y=c_{1}cos(5t)+c_{2}sin(5t)+4tsin(5t)

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y^{''}+25y=40cos(5t),y(0)=y

y^{^{''}} + 25 y = 40 cos (5 t), y (0) = y

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 8 y = e^{2 x} + sin (2 x)

y^{^{''}} + 25 y = sec^{3} (5 t)

y^{^{''}} + 4 y^{^{'}} - 21 y = 5 e^{2 x} + 4

y^{^{''}} + 4 y = 1 - t^{3}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = sin (2 x)

What is the solution for y^{''}+25y=40cos(5t),y(0)=y ?
The solution for y^{''}+25y=40cos(5t),y(0)=y is y=c_{1}cos(5t)+c_{2}sin(5t)+4tsin(5t)