What is the solution for y^{''}-2y^'-8y=7-e^{4x} ?

The solution for y^{''}-2y^'-8y=7-e^{4x} is y=c_{1}e^{4x}+c_{2}e^{-2x}-7/8-(e^{4x}x)/6

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y^{''}-2y^'-8y=7-e^{4x}

y^{^{''}} - 2 y^{^{'}} - 8 y = 7 - e^{4 x}

y = c_{1} e^{4 x} + c_{2} e^{- 2 x} - \frac{7}{8} - \frac{e ^{4 x} x}{6}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 8 y = 7 - e^{4 x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- 2 x} - \frac{7}{8} - \frac{e ^{4 x} x}{6}

y = c_{1} e^{4 x} + c_{2} e^{- 2 x} - \frac{7}{8} - \frac{e ^{4 x} x}{6}

f (x) = x^{3} + 9 x^{2} + 34 x + 15, at x = 7

cos (x^{2})

\frac{d}{d x} (e^{- x^{2}} + x)

g (x) = 3 x, at x = 4

y

What is the solution for y^{''}-2y^'-8y=7-e^{4x} ?
The solution for y^{''}-2y^'-8y=7-e^{4x} is y=c_{1}e^{4x}+c_{2}e^{-2x}-7/8-(e^{4x}x)/6