What is the solution for y^{''}+3y^'+2y=4t ?

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

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y^{''}+3y^'+2y=4t

y^{^{''}} + 3 y^{^{'}} + 2 y = 4 t

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

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = 4 t

Solve linear ODE:

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

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

\int \frac{x}{7 x + 1} d x

d er i v a t i v e f (x) = 9 - 2 x^{2}

\int \frac{1}{2 x + 2 x} d x

\frac{d}{d x} (arccos (x^{2} + y^{2} - 4))

y = c e^{x}

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