What is the solution for y^'e^x+2xy=0 ?

The solution for y^'e^x+2xy=0 is y=e^{-2(-x/(e^x)-1/(e^x))+c_{1}}

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y^'e^x+2xy=0

y^{^{'}} e^{x} + 2 x y = 0

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

Solution steps

y^{'} (x) e^{x} + 2 x y = 0

Solve separable ODE:

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

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

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

(x^{2} - 9) y^{^{'}} + x y = 0

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

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

\frac{d r}{d t} = 0.1872 \cdot r

What is the solution for y^'e^x+2xy=0 ?
The solution for y^'e^x+2xy=0 is y=e^{-2(-x/(e^x)-1/(e^x))+c_{1}}