What is the solution for y^{''}-2y^'-3y=e^{3x} ?

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

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y^{''}-2y^'-3y=e^{3x}

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{3 x}

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

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = e^{3 x}

Solve linear ODE:

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

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

\frac{d ^{2} x}{d t ^{2}} + \frac{d x}{d t} + x = 1, x (0) = 0

y^{^{''}} - y^{^{'}} + y = 14 sin (t)

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 14 y = (x + 1) e^{x}

\frac{d ^{2} y}{d x ^{2}} + 9 y = 3 sin (x)

y^{^{''}} - 7 y^{^{'}} + 12 y = sin (x) - cos (x)

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