What is the solution for y^{''}+2y^'-8y=24e^{2t} ?

The solution for y^{''}+2y^'-8y=24e^{2t} is y=c_{1}e^{2t}+c_{2}e^{-4t}+4te^{2t}

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y^{''}+2y^'-8y=24e^{2t}

y^{^{''}} + 2 y^{^{'}} - 8 y = 24 e^{2 t}

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

Solution steps

y^{''} (t) + 2 y^{'} (t) - 8 y = 24 e^{2 t}

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = x^{3} + x

y^{^{''}} + 10 y^{^{'}} + 16 = 0

y^{^{''}} + 153.06 y^{^{'}} + 12739.6 y = 191094.97

\frac{d ^{3} y}{d x ^{3}} + y = e^{- x} + 1

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 20 y = e^{- 5 x}

What is the solution for y^{''}+2y^'-8y=24e^{2t} ?
The solution for y^{''}+2y^'-8y=24e^{2t} is y=c_{1}e^{2t}+c_{2}e^{-4t}+4te^{2t}