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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = e^{- x} 1 + x

y^{^{''}} - 6 y^{^{'}} + 8 y = 64 x^{2}

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 e^{x}

y^{^{''}} - 2 y^{^{'}} = 4 e^{2 t}

4 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 3 y = cos (2 x)

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