What is the solution for y^{''}-3y^'-4y=e^{3t} ?

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

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y^{''}-3y^'-4y=e^{3t}

y^{^{''}} - 3 y^{^{'}} - 4 y = e^{3 t}

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{1}{4} e^{3 t}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 4 y = e^{3 t}

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{1}{4} e^{3 t}

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{1}{4} e^{3 t}

y = \frac{x ^{3}}{2} + \frac{x ^{2}}{2} - 2 x, y = \frac{x ^{2}}{2}

f (x) = cos^{2} (3 x + 1)

\frac{d}{d x} (ln (4 + x^{2}))

y^{^{'}} - 5 y = 9 e^{8 t}

\int \frac{2}{2 x + 3} d x

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