변수분리형 미분방정식

Rearranging, we have:

$$\begin{aligned} e^y\,\mathrm{d}y &= 2x\,\mathrm{d}x. \\ \text{ Integrating both sides: } \\ \int e^y\,\mathrm{d}y &= \int 2x\,\mathrm{d}x\\ e^y &= x^2+C \\ \text{Since $y=\log2$ when $x=0$, we have} \\ e^{\log 2} &= 0^2+C \\ 2 &= C, \\ \text{and therefore} \\ e^{y} &= x^2+2 \\ y &= \log(x^2+2) \end{aligned}$$

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

$$\begin{aligned} y'&=e^{\frac{y}{3}}\cos t\\ e^{-y/3}\mathrm{d}y&=\cos t\,\mathrm{d}t\\ \int e^{-y/3}\mathrm{d}y&=\int\cos t\,\mathrm{d}t\\ -3e^{-y/3}&=\sin t +C\\ \frac{1}{e^{y/3}}&=\frac{\sin t +C}{-3}\\ e^{y/3} &=\frac{-3}{C+\sin t} \\ \frac{y}{3}&=\log\left(\frac{-3}{C+\sin t}\right) \\y(t)&=3\log\left(\frac{-3}{C+\sin t}\right) \end{aligned}$$

for any constant $C$.

Since the domain of logarithm is $(0,\infty)$, the solution only exists when $C+\sin t<0$.

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

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &= x e^{x^2-\log(y^2)} =\frac{x e^{x^2}}{y^2}\\ y^2\mathrm{d}y&=x e^{x^2}\,\mathrm{d}x\\ \int y^2\mathrm{d}y&=\int x e^{x^2}\,\mathrm{d}x \\ \text{We can guess the antiderivative of $xe^{x^2}$, or use the substitution $u=x^2$, $\mathrm{d}u=2x\mathrm{d}x$.} \\ \frac{y^3}{3}&=\frac{1}{2} e^{x^2}+C'\\ y^3&=\frac{3}{2} e^{x^2}+3C' \\ \text{Since $C'$ can be any constant in $(-\infty,\infty)$, then also $3C'$ can be any constant in $(-\infty,\infty)$, so we replace $3C'$ with the arbitrary constant $C$.} \\ y^3&=\frac{3}{2} e^{x^2}+C \\y&=\sqrt[3]{\frac{3}{2} e^{x^2}+C} \end{aligned}$$

for any constant $C$.

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

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=xe^y\\ \frac{\mathrm{d}y}{e^y}&=x\,\mathrm{d}x\\ \int\frac{\mathrm{d}y}{e^y}&=\int x\,\mathrm{d}x\\ -e^{-y}&=\frac{1}{2} x^2+C\\ e^{-y}&=-\frac{1}{2} x^2-C \\ \text{Since $C$ can be any constant in $(-\infty,\infty)$, then also $-C$ can be any constant in $(-\infty,\infty)$, so we write $C$ instead of $-C$.} \\ e^{-y}&=C-\frac{1}{2} x^2 \\-y&=\log\left(C-\frac{x^2}{2}\right) \\y&=-\log\left(C-\frac{x^2}{2}\right) \end{aligned}$$

for any constant $C$.

The solution only exists for $C-\frac{x^2}{2}>0$. For this to happen, we need $C>0$, and then the domain of the function is those values $x$ for which $|x|<\sqrt{2C}$.

The given differential equation is separable and we solve it accordingly. Cross–multiplying, we rewrite the equation as

$$\begin{aligned} y^2 \frac{\mathrm{d}y}{\mathrm{d}x} &= e^x -2x\\ y^2 \,\mathrm{d}y &= (e^x - 2x ) \,\mathrm{d}x. \\ \text{ Integrating both sides, we find} \\ \int y^2 \,\mathrm{d}y &=\int (e^x - 2x ) \,\mathrm{d}x\\ \frac{1}{3}y^3 &= e^x - x^2 + C \\ \text{Setting $x = 0$ and $y = 3$, we find $\frac{1}{3}3^3=e^0-0^2+C$ and hence $C=8$.} \\ \frac{1}{3}y^3 &= e^x - x^2 + 8\\ y &= (3e^x -3x^2+ 24)^{1/3} \end{aligned}$$

This is a separable differential equation that we solve in the usual way.

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &= -xy^3\\ - \frac{\mathrm{d}y}{y^3} &= x\ \mathrm{d}x\\ \int -\frac{\mathrm{d}y}{y^3} &= \int x\ \mathrm{d}x\\ - \frac{y^{-2}}{-2} &= \frac{x^2}{2}+C\\ y^{-2}&=x^2+2C. \tag{$*$} \\ \text{To have $y=-\frac{1}{4}$ when $x=0$, we must choose $C$ to obey} \\ {\Big(-\frac{1}{4} \Big)}^{-2}& =0 +2C\\ 16&= 2C \\ \text{So, from ($*$),} \\ y^{-2}&=x^2+2C =x^2+16\\ y^2&=\frac{1}{x^2+16} \\ \text{Now, we have two potential candidates for $y(x)$:} \\ y&=\frac{1}{\sqrt{x^2+16}}\qquad\text{OR}\qquad y=-\frac{1}{\sqrt{x^2+16}} \\ \text{We know $y=-\frac14$ when $x=0$. The only function above that fits this is } \\ y&=-\frac{1}{\sqrt{x^2+16}} \end{aligned}$$

So, $f(x) = -\dfrac{1}{\sqrt{x^2+16}}$.

This is a separable differential equation that we solve in the usual way. Cross-multiplying and integrating,

$$\begin{aligned} y\,\mathrm{d}y &= (15x^2 + 4x + 3) \,\mathrm{d}x \\ \int y\,\mathrm{d}y &= \int (15x^2 + 4x + 3) \,\mathrm{d}x \\ \frac{y^2}{2} &= 5x^3 + 2x^2 + 3x + C. \\ \text{Plugging in $x=1$ and $y=4$ gives $ \frac{4^2}2 = 5+2+3+C, $ and so $C=-2$. Therefore} \\ \frac{y^2}{2} &= 5x^3 + 2x^2 + 3x - 2 \\ y^2 &= 10x^3 + 4x^2 + 6x - 4 \\ \text{This leaves us with two possible functions for $y$:} \\ y = \sqrt{10x^3 + 4x^2 + 6x - 4} &\qquad\text{OR}\qquad y = - \sqrt{10x^3 + 4x^2 + 6x - 4} \\ \text{When $x=1$, $y=4$. This only fits the first equation, so} \\ y &= \sqrt{10x^3 + 4x^2 + 6x - 4} \end{aligned}$$

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

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=x^3y \\\frac{\mathrm{d}y}{y}&=x^3\,\mathrm{d}x\\ \int\frac{\mathrm{d}y}{y}&=\int x^3\,\mathrm{d}x\\ \log|y|&=\frac{x^4}{4}+C\\ |y|&=e^{x^4/4+C}=e^{x^4/4}e^C \\ \text{We are told that $y=1$ when $x=0$. That is, $1=e^0e^C$, so $e^C=1$. That is, $C=0$.} \\ |y|&=e^{x^4/4} \\ \text{This leaves us with two potential functions:} \\ y=e^{x^4/4}&\qquad\text{OR}\qquad y=-e^{x^4/4} \\ \text{The first is always positive, and the second is always negative. Since $y=1$ (a positive number) when $x=0$, we see} \\ y&=e^{x^4/4} \end{aligned}$$

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.

The unknown function $f(x)$ satisfies an equation that involves the derivative of $f$. That means we're in differential equation territory. Specifically, we are told that $y=f(x)$ obeys the separable differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=xy$.

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &= xy\\ \frac{\mathrm{d}y}{y} &= x\,\mathrm{d}x \\ \int \frac{\mathrm{d}y}{y} &= \int x\, \mathrm{d}x \\ \log|y| &=\frac{x^2}{2} + C \\ \\ \text{To determine $C$ we set $x=0$ and $y=e$.} \\ \log e &=\frac{0^2}{2}+ C\\ 1&= C \\ \text{So, the solution is } \\ \log|y|&= \frac{x^2}{2} + 1 \\ \text{We are told that $y=f(x)>0$, so may drop the absolute value signs.} \\ \log y &= \frac{x^2}{2}+1 \\ y&=e^{1+\frac{1}{2}x^2}=e\cdot e^{x^2/2} \end{aligned}$$