What is the solution for y^{''}+3y^'+2y=20e^{3t} ?

The solution for y^{''}+3y^'+2y=20e^{3t} is y=c_{1}e^{-t}+c_{2}e^{-2t}+e^{3t}

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y^{''}+3y^'+2y=20e^{3t}

y^{^{''}} + 3 y^{^{'}} + 2 y = 20 e^{3 t}

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + e^{3 t}

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = 20 e^{3 t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + e^{3 t}

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + e^{3 t}

\frac{3}{s ( s ^{3} + 2 s ^{2} + 2 s + 1 )}

\frac{\partial}{\partial y} (e^{\frac{y}{x}} - \frac{e ^{\frac{y}{x}} y}{x})

t + 3 t

n = 0 \sum \infty 2 (\frac{2}{3})^{3 n}

\int \frac{x}{x + 1 - ( x + 1 )} d x

What is the solution for y^{''}+3y^'+2y=20e^{3t} ?
The solution for y^{''}+3y^'+2y=20e^{3t} is y=c_{1}e^{-t}+c_{2}e^{-2t}+e^{3t}