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

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

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y^{''}-y^'=xe^{2x}

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 9 sin (x))

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} + y = 5 e^{t}

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

y^{^{''}} + 4 y = 3, 2 sin (4 t) y (0) = 1

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