What is the solution for y^{''}+3y^'+2y=sin(t) ?

The solution for y^{''}+3y^'+2y=sin(t) is y=c_{1}e^{-t}+c_{2}e^{-2t}+1/10 sin(t)-3/10 cos(t)

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y^{''}+3y^'+2y=sin(t)

y^{^{''}} + 3 y^{^{'}} + 2 y = sin (t)

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{1}{10} sin (t) - \frac{3}{10} cos (t)

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = sin (t)

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{1}{10} sin (t) - \frac{3}{10} cos (t)

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{1}{10} sin (t) - \frac{3}{10} cos (t)

\int (3 - 2 x) sin (5 x) d x

\int cot (9 x) d x

y = x^{5} \cdot e^{x}

\frac{d}{d x} (2 (3^{- x})^{2} + lo g_{2} (x^{4} - 1))

e^{1 - 2 t} + cos^{2} (t)

What is the solution for y^{''}+3y^'+2y=sin(t) ?
The solution for y^{''}+3y^'+2y=sin(t) is y=c_{1}e^{-t}+c_{2}e^{-2t}+1/10 sin(t)-3/10 cos(t)