What is the solution for -y+y^'=x+e^2x ?

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

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-y+y^'=x+e^2x

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'}} + \frac{2}{x} y = x^{2} y^{2}

\frac{d}{d x} (\frac{cos ( x )}{x})

\int - \frac{3}{2 t} d t

\frac{s}{( s - 3 ) ( s + 1 )}

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