What is the solution for y^'-y=xe^{2x}+1 ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (5^{2 x^{3}})

x y^{^{'}} = 4 x^{2} - 2 y, y (1) = 3

y^{2} - 27 x = 0, x + y = 0

f (x) = x^{3} + 4 x - 2, at x = 1

\frac{d}{d x} (\frac{x ^{n + 1}}{n + 1})

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