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

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

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

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

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{1}{6} e^{3 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{1}{6} e^{3 t}

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{1}{6} e^{3 t}

\int (x^{2} - 5) d x

\frac{d}{d x} ((sin (x)) (tan (x)))

\int_{- \infty}^{12} r e^{\frac{r}{4}} d r

y^{^{''}} + 10 y = 0

\int \frac{17 x ^{7}}{315} d x

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