What is the solution for yy^'+xe^{x^2}=0 ?

The solution for yy^'+xe^{x^2}=0 is y=sqrt(-e^{x^2)+c_{1}},y=-sqrt(-e^{x^2)+c_{1}}

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yy^'+xe^{x^2}=0

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

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

Solution steps

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

Solve separable ODE:

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

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

(1 - x^{2}) \frac{d y}{d x} + x y = x, y (0) = 2

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

(y - 1) sin (x) d x - d y = 0

\frac{d y}{d x} = 1 + y^{2}, y (0) = 0

\frac{d y}{d x} + 5 x e^{y} = 0

What is the solution for yy^'+xe^{x^2}=0 ?
The solution for yy^'+xe^{x^2}=0 is y=sqrt(-e^{x^2)+c_{1}},y=-sqrt(-e^{x^2)+c_{1}}