변수분리형 미분방정식

This is a separable differential equation, even if it doesn't quite look like it. First move the $y$ from the left hand side to the right hand side.

$$\begin{aligned} x\frac{\mathrm{d}y}{\mathrm{d}x}+y&=y^2\\ x\frac{\mathrm{d}y}{\mathrm{d}x} &=y^2-y =y(y-1) \\ \frac{\mathrm{d}y}{y(y-1)} &=\frac{\mathrm{d}x}{x} \\ \text{Using the method of partial fractions, we see $\frac{1}{y(y-1)} = \frac{1}{y-1}-\frac{1}{y}.$} \\ \left(\frac{1}{y-1}-\frac{1}{y}\right)\,\mathrm{d}y &=\frac{\mathrm{d}x}{x}\\ \int\left(\frac{1}{y-1}-\frac{1}{y}\right)\,\mathrm{d}y &=\int\frac{\mathrm{d}x}{x}\\ \log|y-1|-\log|y| &=\log |x| + C \\ \log\frac{|y-1|}{|y|}&=\log|x|+C \tag{$*$} \\ \text{To determine $C$ we set $x=1$ and $y=-1$. } \\ \log\frac{|-2|}{|-1|}&=\log|1|+C\\ \log 2&=C \\ \text{Returning to ($*$),} \\ \log\frac{|y-1|}{|y|}&=\log|x|+\log 2\\ \log\left|\frac{y-1}{y}\right|&=\log|2x|\\ \left|\frac{y-1}{y}\right|&=|2x| \\ \text{As $y(1)=-1$ is an initial condition, we have that $x\ge 1$ and $|2x|=2x$. For $x=1$, we have $y=-1$. So at least for $x$ near $1$, we have $y$ near $-1$, so that $\frac{y-1}{y}$ is positive and we may drop the absolute value signs. There remains the possibility that $\frac{y(x)-1}{y(x)}$ changes sign for some larger $x>1$. For now, we will simply ignore that possibility. At the end, we will explicitly check that the $y(x)$ we come up with really does satisfy the differential equation $x\frac{\mathrm{d}y}{\mathrm{d}x}+y=y^2$ and the initial condition $y(1)=-1$.} \\ \frac{y-1}{y}&=2x\\ y-1&=2xy\\ y-2xy&=1\\ y(1-2x)&=1\\ y&=\frac{1}{1-2x} \end{aligned}$$

As a check, we compute:

$$\begin{aligned} x\frac{\mathrm{d}y}{\mathrm{d}x}+y&=x\frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{1}{1-2x}\right\}+y\\ &=x\frac{2}{(1-2x)^2}+\frac{1}{1-2x}\\ &=\frac{2x+(1-2x)}{(1-2x)^2}\\ &=\frac{1}{(1-2x)^2}\\ &=y^2 \\ \text{So, our differential equation is satisfied. Furthermore:} \\ y(1)&=\frac{1}{1-2\times 1}=-1 \end{aligned}$$

as desired. This confirms that our solution is correct.