What is the solution for y^'=xy-xy^2,y(0)=3 ?

The solution for y^'=xy-xy^2,y(0)=3 is y=(9e^{x^2}+6e^{(x^2)/2})/(9e^{x^2)-4}

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y^'=xy-xy^2,y(0)=3

y^{^{'}} = x y - x y^{2}, y (0) = 3

y = \frac{9 e ^{x^{2}} + 6 e ^{\frac{x ^{2}}{2}}}{9 e ^{x^{2}} - 4}

Solution steps

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

Solve separable ODE:

y = \frac{9 e ^{x^{2}} + 6 e ^{\frac{x ^{2}}{2}}}{9 e ^{x^{2}} - 4}

y = \frac{9 e ^{x^{2}} + 6 e ^{\frac{x ^{2}}{2}}}{9 e ^{x^{2}} - 4}

\frac{d y}{d x} = 4 - y^{2}

x (y^{2} - 1) d x - y (x^{2} - 1) d y = 0

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

(2 x - 1) d x + (3 y + 7) d y = 0, y (- 1) = 2

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

What is the solution for y^'=xy-xy^2,y(0)=3 ?
The solution for y^'=xy-xy^2,y(0)=3 is y=(9e^{x^2}+6e^{(x^2)/2})/(9e^{x^2)-4}