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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f^{^{'}} (x) = (x + 1)^{2} \cdot (x - 2)^{3} \cdot (2 x + 3)

\int x cos (x) sin (x) d x

x \to 2 lim (ln (4 - x^{2}))

\int e^{x} + x d x

x \to 0 lim (\frac{2}{x ^{2}})

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