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

The solution for y^{''}+3y^'+2y=3e^{2t} is y=c_{1}e^{-t}+c_{2}e^{-2t}+1/4 e^{2t}

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

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

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{1}{4} e^{2 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{1}{4} e^{2 t}

y = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{1}{4} e^{2 t}

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{- x}

y^{^{'''}} - 3 y^{^{''}} + 2 y^{^{'}} = 12 e^{2 x} + 24 x^{2}

\frac{d ^{2} y}{d t ^{2}} - y = e^{t}, y (0) = 1, y^{^{'}} (0) = 1

a y^{^{''}} - 2 a y^{^{'}} - 3 a y = x^{2}

y^{^{''}} - 5 y^{^{'}} + 6 y = 10 sin (2 x) + 2 e^{3 x}

What is the solution for y^{''}+3y^'+2y=3e^{2t} ?
The solution for y^{''}+3y^'+2y=3e^{2t} is y=c_{1}e^{-t}+c_{2}e^{-2t}+1/4 e^{2t}