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

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

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

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

y = \frac{x ^{2}}{e ^{x^{2}}} + \frac{c _{1}}{e ^{x^{2}}}

Solution steps

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

Solve linear ODE:

y = \frac{x ^{2}}{e ^{x^{2}}} + \frac{c _{1}}{e ^{x^{2}}}

y = \frac{x ^{2}}{e ^{x^{2}}} + \frac{c _{1}}{e ^{x^{2}}}

x \to 0 lim (\frac{x + 1}{x})

\int_{1}^{3} \frac{2}{x ^{3}} d x

\int \frac{x ^{n - 1}}{a + b x ^{n}} d x

\int_{0}^{1} \frac{e ^{- x}}{x} d x

\frac{d y}{d x} = \frac{y ^{2} x}{x ^{2} + 1}

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