Variables Separated Form | Differential Equations

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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