What is the solution for y^{''}+2y^'+5y=2sin(t) ?

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

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y^{''}+2y^'+5y=2sin(t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 2 sin (t)

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 2 sin (t)

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} = 3 x^{2} - 4

y^{^{''}} + 4 y^{^{'}} + 29 = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 x + 3 e^{2 x}

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

y^{^{''}} - 2 y^{^{'}} + y = 3 x^{2} + 6 + 4 e x

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