What is the solution for y^{''}+2y^'-8y=e^{2t}+5t ?

The solution for y^{''}+2y^'-8y=e^{2t}+5t is y=c_{1}e^{2t}+c_{2}e^{-4t}+(e^{2t}t)/6-(5t)/8-5/32

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y^{''}+2y^'-8y=e^{2t}+5t

y^{^{''}} + 2 y^{^{'}} - 8 y = e^{2 t} + 5 t

y = c_{1} e^{2 t} + c_{2} e^{- 4 t} + \frac{e ^{2 t} t}{6} - \frac{5 t}{8} - \frac{5}{32}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 8 y = e^{2 t} + 5 t

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 4 t} + \frac{e ^{2 t} t}{6} - \frac{5 t}{8} - \frac{5}{32}

y = c_{1} e^{2 t} + c_{2} e^{- 4 t} + \frac{e ^{2 t} t}{6} - \frac{5 t}{8} - \frac{5}{32}

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{{- 2 x}} cos (x)

y^{^{''}} - y^{^{'}} - 2 y = 2 e^{x}

(D^{2} - 4 D + 13) y = 24 e^{2 x} \cdot cos (3 x)

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = x e^{x}

y^{^{''}} + 36 y = 11 sec^{2} (6 t)

What is the solution for y^{''}+2y^'-8y=e^{2t}+5t ?
The solution for y^{''}+2y^'-8y=e^{2t}+5t is y=c_{1}e^{2t}+c_{2}e^{-4t}+(e^{2t}t)/6-(5t)/8-5/32