변수분리형 미분방정식

What we're given is an equation relating $y$ to the integral of a function of $y$. What we know how to solve is an equation relating the derivative of $y$ to a function of $y$. We can create this by differentiating the given integral equation. By the Fundamental Theorem of Calculus, part 1:

$$y'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left\{\int_0^x\big(y(t)^2-3y(t)+2\big)\sin t\ \mathrm{d}t\right\} =\big(y(x)^2-3y(x)+2\big)\sin x$$

So $y(x)$ satisfies the differential equation $y'=\big(y^2-3y+2\big)\sin x =(y-2)(y-1)\sin x$ and the initial equation $y(0)=3$ (just substitute $x=0$ into $(*)$). For $y\ne 1,2$:

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=(y-2)(y-1)\sin x\\ \frac{\mathrm{d}y}{(y-2)(y-1)}&=\sin x\ \mathrm{d}x \\\int \frac{\mathrm{d}y}{(y-2)(y-1)}&=\int \sin x\ \mathrm{d}x \\ \text{Using the method of partial fractions, we see $\frac{1}{(y-2)(y-1)}=\frac{1}{y-2}-\frac{1}{y-1}$.} \\ \int\Big[\frac{1}{y-2}-\frac{1}{y-1}\Big]\mathrm{d}y&=\int\sin x\ \mathrm{d}x\\ \log|y-2|-\log|y-1|&=-\cos x+c\\ \log\left|\frac{y-2}{y-1}\right|&=-\cos x+c\\ \left|\frac{y-2}{y-1}\right|&= e^{c-\cos x} \\ \text{The condition $y(0)=3$ forces $\big|\frac{3-2}{3-1}\big|= e^{c-1}$ or $e^c=\frac{1}{2} e$, hence} \\ \left|\frac{y-2}{y-1}\right|&=\frac{1}{2} e^{1-\cos x} \end{aligned}$$

Observe that, when $x=0$, $\frac{y-2}{y-1}=\frac{1}{2}>0$. Furthermore $\frac{1}{2} e^{1-\cos x}$, and hence $\big|\frac{y-2}{y-1}\big|$, can never take the value zero. As $y(x)$ varies continuously with $x$, $y(x)$ must remain larger than 2. Consquently, $\frac{y-2}{y-1}$ remains positive and we may drop the absolute value signs. Hence

$$\begin{aligned} \frac{y-2}{y-1}=\frac{1}{2} e^{1-\cos x} \\ \text{ Solving for $y$,} \\ \frac{y-2}{y-1}&=\frac{1}{2} e^{1-\cos x}\\ 2(y-2)&=e^{1-\cos x}(y-1)\\ 2y-4&=ye^{1-\cos x}-e^{1-\cos x}\\ y\big(2-e^{1-\cos x}\big)&=4-e^{1-\cos x} \\ y&=\frac{4-e^{1-\cos x}}{2-e^{1-\cos x}} \end{aligned}$$

To avoid division by zero in the last step, we need

$$\begin{aligned} e^{1-\cos x}&\neq2\\ 1-\cos x&\neq \log 2\\ \cos x&\neq 1-\log 2 \end{aligned}$$

Let $L=1-\log 2$, for brevity, and note that $L>0$. (This can be seen by observing $2<e$, so, $\log 2<\log e = 1$, hence $1-\log 2>0$.)

We know $x=0$ is in the domain of our function, but the points $x=\pm \arccos(L) = \pm\arccos(1-\log 2)$ are not.

Therefore, the largest interval for which our answer makes sense is $$-\arccos(1-\log 2))>x>\arccos(1-\log 2)$$ or approximately $- 1.259 < x < 1.259$.