# Variables Separated Form

If the equation $\frac{dy}{dx}=ƒ(x,y)$ could be written in the form $\frac{dy}{dx}=\frac{X}{Y}$ where $X$ is a function of $X$ alone and $Y$ is a function of $Y$ alone, it can be written as $Ydy=Xdx$ in which the variables are separated. Then, by integration, we have the general solution, $\int Ydy=\int Xdx+C$ where $C$ is an arbitrary constant of integration. Hence, variables separated form is a differential equation of the form $Ydx=Xdx$ where $Y$ is a function of $y$ alone and $X$ is a function of $x$ alone.

## Solve, by separation of variables, the following differential equations:

### $\frac{dy}{dx}=-\frac{y}{x}$

$\frac{dy}{dx}=-\frac{y}{x}$$\frac{dy}{y}=-\frac{dx}{x}$ Integrating both sides, $\int\frac{dy}{y}=-\frac{dx}{x}+c$ $\log y=-\log x+\log C$ Putting $c=\log C$ to make the simplification easy and nice. Thus, $\log y=\log\frac{C}{x}$ $y=\frac{C}{x}$ $\therefore xy=C$

### $xdx+ydy=0$

$xdx+ydy=0$ $\int(xdx+ydy)=\int 0$ $\frac{x^2}{2}+\frac{y^2}{2}=c$ $x^2+y^2=2c$ $\therefore x^2+y^2=C$ where $C=2c$.

### $(1+x^2)\frac{dy}{dx}=1$

$(1+x^2)\frac{dy}{dx}=1$ $dy=\frac{dx}{1+x^2}$ Integrating both sides, $\int dy=\int\frac{dx}{1+x^2}$ $y=\tan^{-1}+C$

### $\frac{dy}{dx}=\frac{e^x+1}{y}$

$\frac{dy}{dx}=\frac{e^x+1}{y}$ $ydy=(e^x+1)dx$ Integrating both sides, $\int ydy=\int(e^x+1)dx$ $\frac{y^2}{2}=e^x+x+C$ $y^2=2e^x+2x+C$

### $\sqrt{1-x^2}dy+\sqrt{1-y^2}dx=0$

$\sqrt{1-x^2}dy+\sqrt{1-y^2}dx=0$ $\sqrt{1-x^2}dy=-\sqrt{1-y^2}dx$ $\frac{dy}{\sqrt{1-y^2}}=-\frac{dx}{\sqrt{1-x^2}}$ $\frac{dy}{\sqrt{1-y^2}}+\frac{dx}{\sqrt{1-x^2}}=0$ Integrating both sides, $\int\frac{dy}{\sqrt{1-x^2}}+\int\frac{dx}{\sqrt{1-x^2}}=0$ $\sin^{-1}y+\sin^{-1}x=\sin^{-1}C$ $\sin^{-1}x+\sin^{-1}y=\sin^{-1}C$ $\sin^{-1}\left[x\sqrt{1-y^2}+y\sqrt{1-x^2}\right]=\sin^{-1}C$ $\therefore x\sqrt{1-y^2}+y\sqrt{1-x^2}=C$

### $(1+x)ydx+(1+y)xdy=0$

$(1+x)ydx+(1+y)xdy=0$ $(1+x)ydx=-(1+y)xdy$ $\frac{(1+x)}{x}dx=-\frac{(1+y)}{y}dy$ $\left(\frac{1+x}{x}\right)dx+\left(\frac{1+y}{y}\right)dy=0$ $\frac{1}{x}dx+dx+\frac{1}{y}dy+dy=0$ Integrating both sides, we get $\log x+x+\log y+y=C$ $\therefore \log xy+x+y=C$

### $(xy^2+x)dx+(yx^2+y)dy=0$

$(xy^2+x)dx+(yx^2+y)dy=0$ $(y^2+1)xdx=-(x^2+1)ydy$ $\frac{x}{(x^2+1)}dx=-\frac{y}{(y^2+1)}dy$ $\frac{x}{(x^2+1)}dx+\frac{y}{(y^2+1)}dy=0$ $\frac{2x}{(x^2+1)}dx+\frac{2y}{(y^2+1)}dy=0$ Integrating both sides, we get, $\log(x^2+1)+\log(y^2+1)=\log C$ $\log[(x^2+1)(y^2+1)]=\log C$ $\therefore (x^2+1)(y^2+1)=C$

### $\frac{dy}{dx}=e^{x-y}+x^3e^{-y}$

$\frac{dy}{dx}=e^{x-y}+x^3e^{-y}$ $\frac{dy}{dx}=\frac{e^x}{e^y}+\frac{x^3}{e^y}$ $e^ydy=(e^x+x^3)dx$ Integrating bothe sides, we get, $e^y=e^x+\frac{1}{4}x^4+C$

### $\tan xdy+\tan ydx=0$

$\tan xdy+\tan ydx=0$ $\tan xdy=-\tan ydx$ $\frac{dy}{\tan y}=-\frac{dx}{\tan x}$ $\frac{dx}{\tan x}+\frac{dy}{\tan y}=0$ $\frac{\cos x}{\sin x}dx+\frac{\cos y}{\sin y}dy=0$ Integrating both sides, we get $\log(\sin x)+\log(\sin y)=\log C$ $\log(\sin x\sin y)=\log C$ $\therefore \sin x\sin y=C$

### $\frac{dy}{dx}=-\frac{1+\cos 2y}{1-\cos 2x}$

$\frac{dy}{1+\cos 2y}=-\frac{dx}{1-\cos 2x}$ $\frac{dy}{1+\cos 2y}+\frac{dx}{1-\cos 2x}=0$ $\frac{dy}{2\cos^2y}+\frac{dx}{2\sin^2x}=0$ $\sec^2ydy+\operatorname{cosec}^2xdx=0$ Integrating bothe sides, we get, $\tan y-\cot x=C$

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