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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = - \frac{1}{2} (x^{- 2} - x^{4})

\int sin (2 x) \cdot cos (2 x) d x

n = 1 \sum \infty \frac{1}{n} x^{n}

n = 1 \sum \infty (\frac{6}{4})^{n}

\int (1 - x) d x

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