변수분리형 미분방정식

This is a separable differential equation.

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &=\frac{1}{(x^2+x)y}\\ y\,\mathrm{d}y &=\frac{\mathrm{d}x}{x(x+1)} \\ \text{Using partial fractions decomposition, we find $\frac{1}{x(x+1)} = \frac{1}{x}-\frac{1}{x+1}$.} \\ y\,\mathrm{d}y &=\left(\frac{1}{x}-\frac{1}{x+1}\right)\,\mathrm{d}x\\ \int y\,\mathrm{d}y &=\int\left(\frac{1}{x}-\frac{1}{x+1}\right)\,\mathrm{d}x\\ \frac{y^2}{2}&=\log|x|-\log|x+1|+C = \log\left| \frac{x}{x+1}\right|+C\\ \\ \text{To satisfy the initial condition $y(1)=2$ we must choose $C$ to obey} \\ \frac{2^2}{2} &= \log\left|\frac{1}{1+1}\right|+C\\ 2&=\log\frac12+C\\ C&=2-\log\frac{1}{2} \\ \text{So,} \\ \frac{y^2}{2}&= \log\left| \frac{x}{x+1}\right|+2-\log\frac{1}{2}\\ y^2&= 2\log\left| \frac{x}{x+1}\right|+4-2\log\frac{1}{2} \\ \text{Note that the question specifies that $y(1)=2$ is an initial condition. So we always have $x\ge 1$. Then $\frac{x}{x+1}$ is positive, and we can drop the absolute values.} \\ y^2&= 2\log \frac{x}{x+1}+4-2\log\frac{1}{2} \\ \text{This leaves two options for $y(x)$: the positive or negative square root of the right hand side above. Since $y(1)=2$, which is positive, we must choose the positive square root.} \\ y(x)&=\sqrt{2\Big(\log\frac{x}{x+1}-\log\frac{1}{2}+2\Big)}\\ &=\sqrt{4+2\log\frac{2x}{x+1}} \end{aligned}$$

You might worry that $y(x)$ could pass through zero, changing sign, at some $x>1$. But the differential equation says that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{(x^2+x)y}$ is positive whenever $y>0$ and $x\ge 1$. So $y(x)$ is an increasing function whenever $y>0$ and $x\ge 1$. As $y(1)=2$, we have $y(x)\ge 2$ for all $x\ge 1$.