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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 2 y = 4 e^{(- x)} sec^{3} (x)

x, d (x) = x d t (x)

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 3 e^{2 x} - x^{2}

\frac{d ^{2} y}{d x ^{2}} + 4 y = sin (x) cos (x)

y^{^{'''}} + y^{^{'}} = sec (t), y (0) = 3, y^{^{''}} (0) = - 2

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