What is the solution for y^{''}-2y^'-3y=3te^{2t} ?

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

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y^{''}-2y^'-3y=3te^{2t}

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

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

Solution steps

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

Solve linear ODE:

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

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

\int ln (x) \frac{1}{x} d x

(5 x^{6} + 4 x^{3})^{4}

\frac{d}{d x} (x + \frac{16}{x})

f (x) = 24 t^{\frac{3}{2}} + 4 t^{\frac{1}{2}}

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