변수분리형 미분방정식

This is a separable differential equation.

$$\begin{aligned} \frac{1+\sqrt{y^2-4}}{\tan x}\frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\sec x}y\\ y\big[1+\sqrt{y^2-4}\big]\ \mathrm{d}y &= \sec x\ \tan x\ \mathrm{d}x \\ \int y\big[1+\sqrt{y^2-4}\big]\ \mathrm{d}y &= \int \sec x\ \tan x\ \mathrm{d}x \\ \\ \text{For the integral on the left, we use the substitution $u={y^2-4}$, $\frac12\mathrm{d}u=y\,\mathrm{d}y$.} \\ \frac{1}{2}\int \big(1+\sqrt{u}\big)\,\mathrm{d}u&=\sec x +C \\\frac{1}{2} \left(u+\frac{2}{3}u^{3/2}\right)&=\sec x +C \\\frac{1}{2} \left(y^2-4+\frac{2}{3}(y^2-4)^{3/2}\right)&=\sec x +C \\ y^2+\frac{2}{3}(y^2-4)^{3/2}&=2\sec x +2C+4 \\ \text{To find $C$ we set $x=0$ and $y=2$.} \\ 4+\frac{2}{3}\sqrt{4-4}^3&=2\sec (0)+2C+4\\ 4&=2+2C+4\\ 2&=2C+4 \\ \text{So,} \\ y^2+\frac{2}{3}(y^2-4)^{3/2}&=2\sec x +2 \end{aligned}$$

(a) The given differential equation is separable and we solve it accordingly.

$$\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}t}&= k(3-x)(2-x)\\ \frac{\mathrm{d}x}{(x-2)(x-3)}&= k\mathrm{d}t \\ \text{Using the method of partial fractions, we find $\frac{1}{(x-2)(x-3)}=\frac{1}{x-3}-\frac{1}{x-2}$.} \\ \int\Big[\frac{1}{x-3}-\frac{1}{x-2}\Big]\ \mathrm{d}x&= \int k\mathrm{d}t\\ \log|x-3|-\log|x-2|&=kt+C\\ \log\left|\frac{x-3}{x-2}\right|&=kt+C\\ \left|\frac{x-3}{x-2}\right|&=e^{kt+C}=e^{kt}e^C\\ \frac{x-3}{x-2}&=De^{kt} \\ \text{ where $D=\pm e^C$. When $t=0$, $x=1$, forcing} \\ \frac{1-3}{1-2}&=De^{0}\\ D&=2 \\ \text{Hence} \\ \frac{x-3}{x-2}&=2e^{kt}\\ x-3&=2e^{kt}(x-2)\\ x-2e^{kt}x&=3-4e^{kt}\\ x(t)&=\frac{3-4e^{kt}}{1-2e^{kt}} \end{aligned}$$

(b) To evaluate the limit, we could use l'Hopital's rule, but we could also just multiply the numerator and denominator by $e^{-kt}$. Note $\lim\limits_{t \to \infty} e^{-tk}=0$.

$$\begin{aligned} \lim_{t\rightarrow\infty}x(t) &=\lim_{t\rightarrow\infty}\underbrace{\frac{3-4e^{kt}}{1-2e^{kt}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to-\infty}{\mathrm{den}\to-\infty}} =\lim_{t \to \infty} \frac{3-4e^{kt}}{1-2e^{kt}}\cdot\frac{e^{-kt}}{e^{-kt}} =\lim_{t\rightarrow\infty}\frac{3e^{-kt}-4}{e^{-kt}-2} =\frac{0-4}{0-2} =2 \end{aligned}$$

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 $- 1.259 < x < 1.259$.

(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 $y\ne 1,2$,

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

We start by antidifferentiating both sides with respect to $x$.

$$\begin{aligned} \int \left(\frac{\mathrm{d}^2y}{\mathrm{d}{x}^2}\right)\,\mathrm{d}x&=\int\left(\frac{2}{y^3}\cdot\frac{\mathrm{d}y}{\mathrm{d}x}\right)\,\mathrm{d}x \\ \text{The right integral is in exactly the form we would use for a change of variables (substitution) to $y$.} \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=\int\left(\frac{2}{y^3}\right)\,\mathrm{d}y = -\frac{1}{y^2}+C \\ \text{When $y=1$, $\frac{\mathrm{d}y}{\mathrm{d}x}=3$.} \\ 3& =-\frac{1}{1}+C\\ C&=4 \\ \text{So,} \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{1}{y^2}+4 \\ \text{ This is a separable differential equation. } \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{4y^2-1}{y^2} \\ \frac{y^2}{4y^2-1}\,\mathrm{d}y&=\mathrm{d}x\\ \int \frac{y^2}{4y^2-1}\,\mathrm{d}y&=\int \mathrm{d}x\tag{$*$} \end{aligned}$$

We can evaluate the left integral with partial fractions, but because the numerator has the same degree as the denominator, we have to simplify first. We do this by inspection, but you can also use long division.

$$\begin{aligned} \frac{y^2}{4y^2-1}&=\frac{\frac{1}{4}(4y^2-1)+\frac{1}{4}}{4y^2-1}\\ &=\frac{1}{4}\left(1+\frac{1}{4y^2-1}\right)\\ &=\frac{1}{4}\left(1+\frac{1}{(2y-1)(2y+1)}\right)\\ &=\frac{1}{4}\left(1+\frac{1/2}{2y-1}-\frac{1/2}{2y+1}\right) \end{aligned}$$

Now, we return to ($*$).

$$\begin{aligned} \int \mathrm{d}x&=\int \frac{y^2}{4y^2-1}\,\mathrm{d}y\\ &= \int \frac{1}{4}\left(1+\frac{1/2}{2y-1}-\frac{1/2}{2y+1}\right)\,\mathrm{d}y\\ &= \frac{1}{4}\left(y+\frac{1}{4}\log|2y-1| - \frac{1}{4}\log|2y+1|\right)\\ &= \frac{1}{4}\left(y+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right)\\ x+C&=\frac{1}{4}\left(y+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right) \\ \text{When $x=-\frac{1}{16}\log 3$, $y=1$.} \\ -\frac{1}{16}\log 3 +C &=\frac{1}{4}\left(1+\frac14\log\left| \frac{2-1}{2+1}\right|\right) = \frac{1}{4}+\frac{1}{16}\log\frac{1}{3}\\ C&=\frac{1}{4} \\ \text{So,} \\ x+\frac{1}{4}&=\frac{1}{4}\left(y+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right)\\ x&=\frac{1}{4}\left(y-1+\frac{1}{4}\log\left|\frac{2y-1}{2y+1}\right|\right) \end{aligned}$$