# Find field lines of $u = (x^2 , y^2 − y)$ that passes through the point $(1, 1/2)$

Find field lines of $u = (x^2 , y^2 − y)$ that passes through the point $(1, 1/2)$.

So if I've understood it right, you solve the differential equation $$\frac{dx}{x^2} = \frac{dy}{y^2-y}.$$

This then can be written as $(y^2-y)\;dx = x^2\;dy \implies xy^2-xy=xy + C$.
The answer is $y\cdot (e^{1−1/x} + ) = 1$, how did they arrive at that?

$$\frac{dx}{x^2} = \frac{dy}{y^2-y}.$$ It's separable $$\int \frac{dx}{x^2} = \int \frac{dy}{y^2-y}.$$ After integration I got this $$-\frac 1x+K=\ln |\frac {y-1}y|$$ $$\implies \frac {y-1}{y}=Ke^{-\frac 1x}$$ for $(x,y)=(1,1/2) \implies K=-e$

Therefore $$y(1+e^{- \frac 1x+1})=1$$

$(y^2−y)dx=x^2dy\Longrightarrow xy^2−xy=xy+C$

This is the wrong way to go about it. You can't integrate both sides like this because $y$ and $x$ aren't independent. What you should have done is integrate $$\int \frac{dx}{x^2} = \int\frac{dy}{y^2-y},$$ since here the variables are properly separated.

I'll let you integrate this yourself. It's not too hard.

Hint

The solutions of the ODE are given by $$\frac{dx}{x^2} = \frac{dy}{y^2-y}\implies -\dfrac{1}{x}=\log\dfrac{y-1}{y}+c.$$