Find the general integral of…

Find the general integral of $y'=\dfrac{y\varphi'(x)-y^2}{\varphi(x)}$, where $\varphi(x)$ is a given function.

Explanation of Method

Rewrite as $y'-\dfrac{\varphi'(x)}{\varphi(x)}y=-\dfrac{1}{\varphi(x)}y^2$ (Bernoulli, $n=2$) and substitute $v=1/y$; the integrating factor collapses neatly to $\varphi(x)$.

Step 1: Substitute $v=1/y$

Dividing by $y^2$: $-v'-\dfrac{\varphi'(x)}{\varphi(x)}v=-\dfrac{1}{\varphi(x)}$, i.e. $v'+\dfrac{\varphi'(x)}{\varphi(x)}v=\dfrac{1}{\varphi(x)}$.

Step 2: Solve the linear equation

The integrating factor is $\mu(x)=e^{\int\varphi'/\varphi\,dx}=\varphi(x)$, giving $\dfrac{d}{dx}[\varphi(x)v]=1$, so $\varphi(x)v=x+C$.

Result

Final answer:

\[\boxed{\,y=\frac{\varphi(x)}{x+C},\qquad C\in\mathbb{R}}\]