What is the solution for (xy+y^2)y^'=x^2+xy,y(1)=0 ?

The solution for (xy+y^2)y^'=x^2+xy,y(1)=0 is y=sqrt(x^2-1),y=-sqrt(x^2-1)

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(xy+y^2)y^'=x^2+xy,y(1)=0

(x y + y^{2}) y^{^{'}} = x^{2} + x y, y (1) = 0

y = x^{2} - 1, y = - x^{2} - 1

Solution steps

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

Solve separable ODE:

y = x^{2} - 1, y = - x^{2} - 1

y = x^{2} - 1, y = - x^{2} - 1

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

a x^{^{'}} + b x = 0, x (0) = e^{r t}

(4 x^{5} - 16 x^{3} - 2 x) (1 - y^{2}) - 2 y (x^{2} - 4) y^{^{'}} = 0

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

y^{^{'}} = \frac{1 + x}{x y}, y (1) = - 4

What is the solution for (xy+y^2)y^'=x^2+xy,y(1)=0 ?
The solution for (xy+y^2)y^'=x^2+xy,y(1)=0 is y=sqrt(x^2-1),y=-sqrt(x^2-1)