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

The solution for y^{''}+3y^'+2y=e^t is y=c_{1}e^{-t}+c_{2}e^{-2t}+1/6 e^t

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y^{''}+3y^'+2y=e^t

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{t}

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{1}{6} e^{t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{1}{6} e^{t}

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{1}{6} e^{t}

y^{^{''}} - y = x sin (x)

y^{^{''}} + 4 y = - 12 sin (2 x)

y^{^{'''}} - 4 y^{^{''}} = x + x e^{4 x}

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = x e^{x} - e^{- x} + 7

y^{^{''}} + 36 y = csc (6 x)

What is the solution for y^{''}+3y^'+2y=e^t ?
The solution for y^{''}+3y^'+2y=e^t is y=c_{1}e^{-t}+c_{2}e^{-2t}+1/6 e^t