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

The solution for y^{''}-3y^'-4y=e^{4t} is y=c_{1}e^{4t}+c_{2}e^{-t}+(e^{4t}t)/5

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

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

y = c_{1} e^{4 t} + c_{2} e^{- t} + \frac{e ^{4 t} t}{5}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- t} + \frac{e ^{4 t} t}{5}

y = c_{1} e^{4 t} + c_{2} e^{- t} + \frac{e ^{4 t} t}{5}

\int (x, π x) d x

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

(x^{2} sin (x^{2}))^{^{'}}

\int \frac{( 4 x ^{3} - 6 x ^{4} )}{2 x ^{6}} d x

f (x) = 2 x^{4} + 3 x - 3, (2, 35)

What is the solution for y^{''}-3y^'-4y=e^{4t} ?
The solution for y^{''}-3y^'-4y=e^{4t} is y=c_{1}e^{4t}+c_{2}e^{-t}+(e^{4t}t)/5