What is the solution for y^{''}+6y^'+25y=15cos(5t) ?

The solution for y^{''}+6y^'+25y=15cos(5t) is y=e^{-3t}(c_{1}cos(4t)+c_{2}sin(4t))+1/2 sin(5t)

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y^{''}+6y^'+25y=15cos(5t)

y^{^{''}} + 6 y^{^{'}} + 25 y = 15 cos (5 t)

y = e^{- 3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{1}{2} sin (5 t)

Solution steps

y^{''} (t) + 6 y^{'} (t) + 25 y = 15 cos (5 t)

Solve linear ODE:

y = e^{- 3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{1}{2} sin (5 t)

y = e^{- 3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{1}{2} sin (5 t)

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 t^{6} e^{2 t}

y^{^{''}} - 2 y^{^{'}} + y = t^{- 7} e^{t}

y^{^{''}} - y - 2 = 0, y (0) = 3

y^{^{''}} - 6 y^{^{'}} + 8 y = 2

y^{^{''}} + 3 y^{^{'}} - 4 = 0

What is the solution for y^{''}+6y^'+25y=15cos(5t) ?
The solution for y^{''}+6y^'+25y=15cos(5t) is y=e^{-3t}(c_{1}cos(4t)+c_{2}sin(4t))+1/2 sin(5t)