What is the solution for y^{''}+2y^'-8y=2e^{4t} ?

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

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y^{''}+2y^'-8y=2e^{4t}

y^{^{''}} + 2 y^{^{'}} - 8 y = 2 e^{4 t}

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

Solution steps

y^{''} (t) + 2 y^{'} (t) - 8 y = 2 e^{4 t}

Solve linear ODE:

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

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

d (x) = \frac{3}{2 x}

(d^{2} + 6 d + 9) y = \frac{e ^{- 3} x}{x ^{3}}

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

x^{^{''}} + 16 x = 36 cos (2 t)

y^{^{''}} - 12 y^{^{'}} + 36 y = 7 e^{6 x}

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