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

The solution for y^{''}-y-12y=e^{4x} is y=c_{1}e^{sqrt(13)x}+c_{2}e^{-sqrt(13)x}+1/3 e^{4x}

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

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

y = c_{1} e^{13 x} + c_{2} e^{- 13 x} + \frac{1}{3} e^{4 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{13 x} + c_{2} e^{- 13 x} + \frac{1}{3} e^{4 x}

y = c_{1} e^{13 x} + c_{2} e^{- 13 x} + \frac{1}{3} e^{4 x}

\frac{d ^{2} y}{d x ^{2}} - y = \frac{1}{x}

6 y^{^{''}} - y^{^{'}} - 2 y = 3 sin (10) x

d, d (x) = \frac{3 x - 2}{x + 1}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 x^{2} - 4 x + 2

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} = 37 sin (x)

What is the solution for y^{''}-y-12y=e^{4x} ?
The solution for y^{''}-y-12y=e^{4x} is y=c_{1}e^{sqrt(13)x}+c_{2}e^{-sqrt(13)x}+1/3 e^{4x}