변수분리형 미분방정식

(a) Setting $x=0$ gives

$$\begin{aligned} f(0)=3+\int_0^0\big(f(t)-1\big)\big(f(t)-2\big)\ \mathrm{d}t= 3 \end{aligned}$$

(b) By the Fundamental Theorem of Calculus part 1,

$$\begin{aligned} f'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\int_0^x\big(f(t)-1\big)\big(f(t)-2\big)\ \mathrm{d}t =\big(f(x)-1\big)\big(f(x)-2\big) \end{aligned}$$

Thus $y=f(x)$ obeys the differential equation $y'=(y-1)(y-2)$.

(c) If $y\ne 1,2$,

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=(y-1)(y-2) \\ \frac{\mathrm{d}y}{(y-1)(y-2)}&=\mathrm{d}x\\ \int\frac{\mathrm{d}y}{(y-1)(y-2)}&=\int\mathrm{d}x \\ \text{Using the method of partial fractions,} \\ \int\left(\frac{1}{y-2}-\frac{1}{y-1}\right)\mathrm{d}y&=\int \mathrm{d}x \\ \log|y-2|-\log|y-1|&=x+C\\ \log\left|\frac{y-2}{y-1}\right|&=x+C \\ \text{ Observe that $\frac{\mathrm{d}y}{\mathrm{d}x}=(y-1)(y-2)>0$ for all $y\ge 2$. That is, $f(x)$ is increasing at all $x$ for which $f(x)>2$. As $f(0)=3$, $f(x)$ increases for all $x\ge 0$, and $f(x)\ge 3$ for all $x\ge 0$. So we may drop the absolute value signs.} \\ \log\frac{f(x)-2}{f(x)-1}&=x+C\\ \frac{f(x)-2}{f(x)-1}&=e^Ce^x \\ \text{ At $x=0$, $\frac{f(x)-2}{f(x)-1}=\frac{1}{2}$ so $e^C=\frac{1}{2}$.} \\ \frac{f(x)-2}{f(x)-1}&=\frac{1}{2} e^x\\ 2f(x)-4&=[f(x)-1]e^x\\ [2-e^x]f(x)&=4-e^x \\ f(x)&=\frac{4-e^x}{2-e^x} \end{aligned}$$