What is the solution for dy+(2xy-4e^{-x^2})dx=0 ?

The solution for dy+(2xy-4e^{-x^2})dx=0 is y=(4x)/(e^{x^2)}+(c_{1})/(e^{x^2)}

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dy+(2xy-4e^{-x^2})dx=0

d y + (2 x y - 4 e^{- x^{2}}) d x = 0

y = \frac{4 x}{e ^{x^{2}}} + \frac{c _{1}}{e ^{x^{2}}}

Solution steps

d y + (2 x y - 4 e^{- x^{2}}) d x = 0

Solve linear ODE:

y = \frac{4 x}{e ^{x^{2}}} + \frac{c _{1}}{e ^{x^{2}}}

y = \frac{4 x}{e ^{x^{2}}} + \frac{c _{1}}{e ^{x^{2}}}

(4, 1), (3, 1)

(x + y)^{2} d x + (2 x y + x^{2} - 6) d y = 0

(x lo g_{2} (x))^{2}

\int \frac{lo g _{10} ( x )}{x} d x

\frac{d}{d t} (e^{t^{2} - t})

What is the solution for dy+(2xy-4e^{-x^2})dx=0 ?
The solution for dy+(2xy-4e^{-x^2})dx=0 is y=(4x)/(e^{x^2)}+(c_{1})/(e^{x^2)}