What is the solution for y^{''}-y^'-12y=e^{4x} ?

The solution for y^{''}-y^'-12y=e^{4x} is y=c_{1}e^{4x}+c_{2}e^{-3x}+(e^{4x}x)/7

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y^{''}-y^'-12y=e^{4x}

y^{^{''}} - y^{^{'}} - 12 y = e^{4 x}

y = c_{1} e^{4 x} + c_{2} e^{- 3 x} + \frac{e ^{4 x} x}{7}

Solution steps

y^{''} (x) - y^{'} (x) - 12 y = e^{4 x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- 3 x} + \frac{e ^{4 x} x}{7}

y = c_{1} e^{4 x} + c_{2} e^{- 3 x} + \frac{e ^{4 x} x}{7}

\int \frac{sin ( x ) tan ( x )}{1} d x

\int_{0}^{4} \frac{1}{16 - x ^{2}} d x

x \to 0 + lim (x^{2 x})

\frac{\partial}{\partial t} (e^{t})

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What is the solution for y^{''}-y^'-12y=e^{4x} ?
The solution for y^{''}-y^'-12y=e^{4x} is y=c_{1}e^{4x}+c_{2}e^{-3x}+(e^{4x}x)/7