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)

2 s

(2 x + y) d x - x d y = 0

\frac{\partial}{\partial y} (x + y)

(x^{2} y^{4} + x^{6}) d x - x^{3} y^{3} d y = 0

t y^{^{''}} - y^{^{'}} = 0

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)