Separation Of Variables Method

I have discussed Separation of variables method of 1st order derivative in the previous article. Here is the brief introduction.
If the given equation, say Mdx + Ndy = 0, where M and N are functions of x and y, can be put in the form
\[{{f}_{1}}(x)dx+{{f}_{2}}dy=0\]
Then we say that the variables are separable. Such equations can be solved by integrating both sides.
It is not always possible to separate the variables x and y from each other. For this we require certain substitutions which will serve our purpose.
You can read my article on Cheat Sheet of First Order First Degree Differential Equation Solution Methods for better understanding.

Below are 21 most important separable variable questions which I am solving for you.

\[Solve:{{\log }_{e}}\left( \frac{dy}{dx} \right)=ax+by\]
Solution:
\[Given,{{\log }_{e}}\left( \frac{dy}{dx} \right)=ax+by\]
\[\therefore \frac{dy}{dx}={{e}^{ax+by}}\Rightarrow \frac{dy}{dx}={{e}^{ax}}.{{e}^{by}}\]
\[\Rightarrow {{e}^{-by}}dy={{e}^{ax}}dx\]
Integrating we get,
\[\int{{{e}^{-by}}dy}=\int{{{e}^{ax}}dx+c}\]
\[\Rightarrow \frac{{{e}^{-by}}}{-b}=\frac{{{e}^{ax}}}{a}+c\]
\[\Rightarrow b{{e}^{ax}}+a{{e}^{-by}}+abc=0\]
where c is arbitrary constant.

\[Solve:x\left( 1+{{y}^{2}} \right)dx+y\left( 1+{{x}^{2}} \right)dy=0\]
Solution:
\[Given,x\left( 1+{{y}^{2}} \right)dx+y\left( 1+{{x}^{2}} \right)dy=0\]
\[\Rightarrow \frac{xdx}{\left( 1+{{x}^{2}} \right)}+\frac{ydy}{\left( 1+{{y}^{2}} \right)}=0\]
\[\Rightarrow \frac{1}{2}\frac{d\left( 1+{{x}^{2}} \right)}{\left( 1+{{x}^{2}} \right)}+\frac{1}{2}\frac{d\left( 1+{{y}^{2}} \right)}{\left( 1+{{y}^{2}} \right)}=0\]
Integrating we get,
\[\frac{1}{2}\int{\frac{d\left( 1+{{x}^{2}} \right)}{\left( 1+{{x}^{2}} \right)}+}\frac{1}{2}\int{\frac{d\left( 1+{{y}^{2}} \right)}{\left( 1+{{y}^{2}} \right)}=\frac{1}{2}\log c}\]
\[\Rightarrow \frac{1}{2}\log \left( 1+{{x}^{2}} \right)+\frac{1}{2}\log \left( 1+{{y}^{2}} \right)=\frac{1}{2}\log c\]
\[\Rightarrow \log \left\{ \left( 1+{{x}^{2}} \right)\left( 1+{{y}^{2}} \right) \right\}=\log c\]
\[\therefore \left( 1+{{x}^{2}} \right)\left( 1+{{y}^{2}} \right)=c\]
where c is arbitrary constant.

\[Solve:3{{e}^{x}}\tan ydx+\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=0\]
Solution:
\[We-have,3{{e}^{x}}\tan ydx+\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=0\]
\[\Rightarrow \frac{3{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx=-\frac{{{\sec }^{2}}y}{\tan y}dy\]
Integrating we get,
\[\int{\frac{3{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx}=-\int{\frac{{{\sec }^{2}}y}{\tan y}dy-\log c}\]
\[\Rightarrow 3\int{\frac{d\left( 1-{{e}^{x}} \right)}{\left( 1-{{e}^{x}} \right)}dx}=\int{\frac{{{\sec }^{2}}y}{\tan y}dy+\log c}\]
\[\Rightarrow 3\log \left( 1-{{e}^{x}} \right)=\log \tan y+\log c\]
\[\Rightarrow \log {{\left( 1-{{e}^{x}} \right)}^{3}}=\log \left( c\tan y \right)\]
\[\therefore {{\left( 1-{{e}^{x}} \right)}^{3}}=c\tan y\]
where c is arbitrary constant.

\[Solve:x\sqrt{1-{{y}^{2}}}dx+y\sqrt{1-{{x}^{2}}}dy=0\]
Solution:
\[Given,x\sqrt{1-{{y}^{2}}}dx+y\sqrt{1-{{x}^{2}}}dy=0\]
\[\Rightarrow \frac{xdx}{\sqrt{1-{{x}^{2}}}}=-\frac{ydy}{\sqrt{1-{{y}^{2}}}}\]
Integrating we get,
\[\Rightarrow \int{\frac{xdx}{\sqrt{1-{{x}^{2}}}}}=-\int{\frac{ydy}{\sqrt{1-{{y}^{2}}}}+c}\]
\[Let,1-{{x}^{2}}={{u}^{2}},1-{{y}^{2}}={{v}^{2}}\]
\[\Rightarrow -2xdx=2udy,-2ydy=2vdv\]
\[\Rightarrow xdx=-udy,ydy=-vdv\]

\[Solve:\sin x\frac{dy}{dx}-y\cos x+{{y}^{2}}=0\]
Solution:
\[Given,\sin x\frac{dy}{dx}-y\cos x+{{y}^{2}}=0\]
\[\Rightarrow \frac{1}{{{y}^{2}}}\frac{dy}{dx}.\sin x+\left( -\frac{1}{y} \right)\cos x+1=0\]
\[\Rightarrow \sin x.\frac{d}{dx}\left( -\frac{1}{y} \right)+\left( -\frac{1}{y} \right).\frac{d}{dx}\left( \sin x \right)+1=0\]
\[\Rightarrow \frac{d}{dx}\left( -\frac{\sin x}{y} \right)=-1\]
\[\Rightarrow d\left( -\frac{\sin x}{y} \right)=-dx\]
Integrating we get,
\[\int{d\left( -\frac{\sin x}{y} \right)=-\int{dx-c}}\]
\[\Rightarrow -\frac{\sin x}{y}=-x-c\]
\[\therefore \sin x=y\left( x+c \right)\]
where c is arbitrary constant.

\[Solve:\frac{dy}{dx}=\frac{{{y}^{2}}+y+1}{{{x}^{2}}+x+1}\]
Solution:
\[Given,\frac{dy}{dx}=\frac{{{y}^{2}}+y+1}{{{x}^{2}}+x+1}\]
\[\Rightarrow \frac{dy}{{{y}^{2}}+y+1}=\frac{dx}{{{x}^{2}}+x+1}\]
\[\Rightarrow \frac{dy}{{{y}^{2}}+2.y.\frac{1}{2}+\frac{1}{4}+1-\frac{1}{4}}=\frac{dx}{{{x}^{2}}+2.x.\frac{1}{2}+\frac{1}{4}+1-\frac{1}{4}}\]
\[\Rightarrow \frac{dy}{{{\left( y+\frac{1}{2} \right)}^{2}}-{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}=\frac{dx}{{{\left( x+\frac{1}{2} \right)}^{2}}-{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}\]
Integrating we get,
\[\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left[ \frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right]=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left[ \frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right]+\frac{2}{\sqrt{3}}{{\tan }^{-1}}\frac{3c}{\sqrt{3}}\]
\[\Rightarrow {{\tan }^{-1}}\frac{2y+1}{\sqrt{3}}-{{\tan }^{-1}}\frac{2x+1}{\sqrt{{}}}={{\tan }^{-1}}\frac{3c}{\sqrt{3}}\]
\[\Rightarrow {{\tan }^{-1}}\frac{\frac{2y+1}{\sqrt{3}}-\frac{2x+1}{\sqrt{3}}}{1+\frac{2y+1}{\sqrt{3}}.\frac{2x+1}{\sqrt{3}}}={{\tan }^{-1}}\frac{3c}{\sqrt{3}}\]
\[\Rightarrow \frac{\sqrt{3}\left( 2y+1-2x-1 \right)}{3+\left( 2y+1 \right)\left( 2x+1 \right)}=\sqrt{3}c\]
\[\Rightarrow \frac{2\left( y-x \right)}{3+4xy+2y+2x+1}=c\]
\[\Rightarrow \frac{2\left( y-x \right)}{2\left( 2+2xy+x+y \right)}=c\]
\[\therefore y-x=c(x+y+2xy+2)\]

\[Solve:y-x\frac{dy}{dx}=a\left( {{y}^{2}}+\frac{dy}{dx} \right)\]
Solution:
\[Given,y-x\frac{dy}{dx}=a\left( {{y}^{2}}+\frac{dy}{dx} \right)\]
\[\Rightarrow y-a{{y}^{2}}=x\frac{dy}{dx}+a\frac{dy}{dx}\]
\[\Rightarrow y\left( 1-ay \right)=\left( x+a \right)\frac{dy}{dx}\]
\[\Rightarrow \frac{dx}{\left( x+a \right)}=\frac{dy}{y\left( 1-ay \right)}\]
\[\Rightarrow \frac{dx}{\left( x+a \right)}=\frac{ay+1-ay}{y\left( 1-ay \right)}dy\]
\[\Rightarrow \frac{dx}{\left( x+a \right)}=\left[ \frac{a}{\left( 1-ay \right)}+\frac{1}{y} \right]dy\]
Integrating we get,
\[\Rightarrow \int{\frac{dx}{\left( x+a \right)}}=\int{\left[ \frac{a}{\left( 1-ay \right)}+\frac{1}{y} \right]dy+\log c}\]
\[\Rightarrow \log \left( x+a \right)=-\log \left( 1-ay \right)+\log y+\log c\]
\[\Rightarrow \log \left( x+a \right)+\log \left( 1-ay \right)=\log y+\log c\]
\[\Rightarrow \log \left\{ \left( x+a \right)\left( 1-ay \right) \right\}=\log cy\]
\[\therefore \left( x+a \right)\left( 1-ay \right)=cy\]
c being an arbitrary constant.

\[Solve:{{\left( x+y+1 \right)}^{2}}\frac{dy}{dx}=1\]
Solution:
\[Given,{{\left( x+y+1 \right)}^{2}}\frac{dy}{dx}=1……….(1)\]
\[Putting,x+y+1=z\Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx}\]
\[\therefore \frac{dy}{dx}=\frac{dz}{dx}-1\]
Therefore the equation (1) becomes,
\[{{z}^{2}}\left( \frac{dz}{dx}-1 \right)=1\]
\[\Rightarrow \frac{dz}{dx}=\frac{1+{{z}^{2}}}{{{z}^{2}}}\]
\[\Rightarrow \frac{{{z}^{2}}}{1+{{z}^{2}}}dz=dx\Rightarrow \frac{1+{{z}^{2}}-1}{1+{{z}^{2}}}dz=dx\]
Integrating we get,
\[\int{\left( 1-\frac{1}{1+{{z}^{2}}} \right)}dz=\int{dx}+c\]
\[\Rightarrow z-{{\tan }^{-1}}z=x+c\]
\[\Rightarrow x+y+1-{{\tan }^{-1}}\left( x+y+1 \right)=x+c\]
\[\therefore y={{\tan }^{-1}}\left( x+y+1 \right)+k;[k=c-1]\]

\[Solve:\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{{{a}^{2}}-{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}}\]
Solution:
\[Given,\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{{{a}^{2}}-{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}}\]
\[Put,x=r\cos \theta ,y=r\sin \theta \]
\[such-that,{{x}^{2}}+{{y}^{2}}={{r}^{2}}\]
\[\therefore 2xdx+2ydy=2rdr\]
\[\Rightarrow xdx+ydy=rdr\]
\[And,\tan \theta =\frac{y}{x}\Rightarrow \theta ={{\tan }^{-1}}\frac{y}{x}\]
\[\therefore {{\sec }^{2}}\theta d\theta =\frac{xdy-ydx}{{{x}^{2}}}\]
\[\Rightarrow \frac{{{r}^{2}}}{{{x}^{2}}}d\theta =\frac{xdy-ydx}{{{x}^{2}}},\left[ \because \sec \theta =\frac{r}{x} \right]\]
\[\therefore {{r}^{2}}d\theta =xdy-ydx\]
Thus the given equation becomes,
\[\frac{rdr}{{{r}^{2}}d\theta }=\sqrt{\frac{{{a}^{2}}-{{r}^{2}}}{{{r}^{2}}}}\]
\[\Rightarrow \frac{dr}{\sqrt{{{a}^{2}}-{{r}^{2}}}}=d\theta \]
Integrating we get,
\[\int{\frac{dr}{\sqrt{{{a}^{2}}-{{r}^{2}}}}}=\int{d\theta +c}\]
\[\Rightarrow {{\sin }^{-1}}\frac{r}{a}=\theta +c\]
\[\Rightarrow r=a\sin (\theta +c)\]
\[\therefore \sqrt{{{x}^{2}}+{{y}^{2}}}=a\sin \left( {{\tan }^{-1}}\frac{y}{x}+c \right)\]
c being an arbitrary constant.

\[Solve:{{x}^{2}}\left( xdx+ydy \right)+2y\left( xdy-ydx \right)=0\]
Solution:
\[Given,{{x}^{2}}\left( xdx+ydy \right)+2y\left( xdy-ydx \right)=0\]
\[Put,x=r\cos \theta ,y=r\sin \theta \]
\[such-that,{{x}^{2}}+{{y}^{2}}={{r}^{2}}\]
\[\therefore 2xdx+2ydy=2rdr\]
\[\Rightarrow xdx+ydy=rdr\]
\[And,\tan \theta =\frac{y}{x}\Rightarrow \theta ={{\tan }^{-1}}\frac{y}{x}\]
\[\therefore {{\sec }^{2}}\theta d\theta =\frac{xdy-ydx}{{{x}^{2}}}\]
\[\Rightarrow \frac{{{r}^{2}}}{{{x}^{2}}}d\theta =\frac{xdy-ydx}{{{x}^{2}}},\left[ \because \sec \theta =\frac{r}{x} \right]\]
\[\therefore {{r}^{2}}d\theta =xdy-ydx\]
Thus the given equation becomes,
\[{{r}^{2}}{{\cos }^{2}}\theta .rdr+2r\sin \theta .{{r}^{2}}d\theta =0\]
\[\Rightarrow dr+\frac{2\sin \theta }{{{\cos }^{2}}\theta }d\theta =0,\left[ \because r\ne 0 \right]\]
Integrating we get,
\[\int{dr}+\int{\frac{2\sin \theta }{{{\cos }^{2}}\theta }d\theta }=c\]
\[\Rightarrow r+\frac{2}{\cos \theta }=c\]
\[\therefore \sqrt{{{x}^{2}}+{{y}^{2}}}+\frac{2r}{r\cos \theta }=c\]
\[\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}+\frac{2\sqrt{{{x}^{2}}+{{y}^{2}}}}{x}=c\]
\[\therefore \sqrt{{{x}^{2}}+{{y}^{2}}}\left( x+2 \right)=cx\]

\[Solve:\frac{dy}{dx}=x{{e}^{y-{{x}^{2}}}},if,y=0,when,x=0\]
Solution:
\[Given,\frac{dy}{dx}=x{{e}^{y-{{x}^{2}}}}\]
\[\Rightarrow {{e}^{-y}}dy=x{{e}^{-{{x}^{2}}}}dx\]
Integrating we get,
\[\int{{{e}^{-y}}dy}=\int{x{{e}^{-{{x}^{2}}}}dx}+c\]
\[\Rightarrow -{{e}^{-y}}=\frac{1}{2}\int{{{e}^{-{{x}^{2}}}}d\left( {{x}^{2}} \right)+c}\]
\[\Rightarrow -{{e}^{-y}}=-\frac{1}{2}{{e}^{-{{x}^{2}}}}+c\]
\[Given,y=0,when,x=0\]
\[\therefore -1=-\frac{1}{2}+c\Rightarrow c=-\frac{1}{2}\]
Thus the particular solution is
\[-{{e}^{-y}}=-\frac{1}{2}{{e}^{-{{x}^{2}}}}-\frac{1}{2}\]
\[\therefore 2{{e}^{-y}}={{e}^{-{{x}^{2}}}}+1\]

\[Solve:\frac{dy}{dx}=\frac{\cos \left( \log x \right)}{\log y}\]
Solution:
\[Given,\frac{dy}{dx}=\frac{\cos \left( \log x \right)}{\log y}\]
\[\Rightarrow \log ydy=\cos \left( \log x \right)dx\]
Integrating we get,
\[\int{\log ydy}=\int{\cos \left( \log x \right)dx+c}\]
\[Let,\log x=z,i.e.,x={{e}^{z}}\]
\[\therefore \frac{1}{x}dx=dz\Rightarrow dx={{e}^{z}}dz\]
\[\therefore \int{\log ydy}=\int{{{e}^{z}}\cos zdz+c}\]
\[\Rightarrow {{I}_{1}}={{I}_{2}}+c……….(1)\]
\[\therefore {{I}_{1}}=\int{\log ydy}=\log y\int{dy-\int{\left\{ \frac{d}{dy}\log y.\int{dy} \right\}}}dy\]
\[\Rightarrow {{I}_{1}}=y\log y-\int{\frac{1}{y}.ydy=y\log y-y}\]
\[\therefore {{I}_{2}}=\int{{{e}^{z}}\cos zdz=\cos z\int{{{e}^{z}}dz-\int{\left\{ \frac{d}{dz}\cos z\int{{{e}^{z}}dz} \right\}}}}dz\]
\[\Rightarrow {{I}_{2}}={{e}^{z}}\cos z+\int{{{e}^{z}}\sin zdz}\]
\[\Rightarrow {{I}_{2}}={{e}^{z}}\cos z+\sin z\int{{{e}^{z}}dz-\int{\left\{ \frac{d}{dz}\sin z\int{{{e}^{z}}dz} \right\}}dz}\]
\[\Rightarrow {{I}_{2}}={{e}^{z}}\cos z+{{e}^{z}}\sin z-\int{{{e}^{z}}\cos zdz}\]
\[\Rightarrow {{I}_{2}}={{e}^{z}}\cos z+{{e}^{z}}\sin z-{{I}_{2}}\]
\[\therefore {{I}_{2}}=\frac{1}{2}{{e}^{z}}\left( \cos z+\sin z \right)\]
Thus equation (1) becomes,
\[y\log y-y=\frac{1}{2}{{e}^{z}}\left( \cos z+\sin z \right)+c\]
\[\therefore \left( y-1 \right)\log y=\frac{x}{2}\left\{ \cos \left( \log x \right)+\sin \left( \log x \right) \right\}+c\]

\[Solve:{{\tan }^{-1}}\left( \frac{dy}{dx} \right)=x+y\]
Solution:
\[Given,{{\tan }^{-1}}\left( \frac{dy}{dx} \right)=x+y\]
\[\Rightarrow \frac{dy}{dx}=\tan \left( x+y \right)\]
\[Let,x+y=z\Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx}\]
Therefore the given equation becomes,
\[\frac{dz}{dx}-1=\tan z\]
\[\frac{dz}{dx}=1+\tan z=1+\frac{\sin z}{\cos z}\]
\[\Rightarrow \frac{dz}{dx}=\frac{\cos z+\sin z}{\cos z}\]
\[\Rightarrow \frac{\cos z}{\cos z+\sin z}dz=dx\]
Integrating we get,
\[\int{\frac{\cos z}{\cos z+\sin z}dz}=\int{dx+c}\]
\[\Rightarrow \frac{1}{2}\int{\frac{\left( \cos z+\sin z \right)+\left( \cos z-\sin z \right)}{\cos z+\sin z}}dz=\int{dx+c}\]
\[\Rightarrow \frac{1}{2}\int{dz+\frac{1}{2}\int{\frac{d\left( \sin z+\cos z \right)}{\sin z+\cos z}}}=\int{dx+c}\]
\[\Rightarrow \frac{1}{2}z+\frac{1}{2}\log \left( \sin z+\cos z \right)=x+c\]
\[\Rightarrow \left( x+y \right)+\log \left\{ \sin \left( x+y \right)+\cos \left( x+y \right) \right\}=2x+2c\]
\[\therefore y-x+\log \left\{ \sin \left( x+y \right)+\cos \left( x+y \right) \right\}=k,\left[ where,k=2c \right]\]

\[Solve:\left\{ 1+xy\cos \left( xy \right) \right\}dx+{{x}^{2}}\cos \left( xy \right)dy=0\]
Solution:
The given equation can be written as
\[\left\{ \frac{1+xy\cos \left( xy \right)}{xy} \right\}ydx+\left\{ \cos \left( xy \right) \right\}xdy=0\]
Which is of the form,
\[{{f}_{1}}(xy)ydx+{{f}_{2}}(xy)xdy=0\]
\[Put,xy=u\Rightarrow ydx+xdy=du\]
\[\Rightarrow xdy=du-\frac{u}{x}dx\]
The given equation becomes,
\[\frac{1+u\cos u}{u}.\frac{u}{x}dx+\cos u\left( du-\frac{u}{x}dx \right)=0\]
\[\Rightarrow \frac{dx}{x}+\frac{u}{x}\cos udx+\cos udu-\frac{u}{x}\cos udx=0\]
\[\Rightarrow \frac{dx}{x}+\cos udu=0\]
Integrating we get,
\[\int{\frac{dx}{x}}+\int{\cos udu}=\log c\]
\[\Rightarrow \log x+\sin u=\log c\]
\[\Rightarrow \log \frac{x}{c}=-\sin u\]
\[\therefore x=c{{e}^{-\sin u}}\Rightarrow x=c{{e}^{-\sin \left( xy \right)}}\]

\[If,f'(x)+f(x)=0,and,f(0)=2,find,f(x)\]
Solution:
\[Given,f'(x)+f(x)=0\]
\[Let,f(x)=z\Rightarrow f'(x)=\frac{dz}{dx}\]
Therefore the given equation becomes,
\[\frac{dz}{dx}+z=0\]
\[\Rightarrow \frac{dz}{z}=-dx\]
Integrating we get,
\[\int{\frac{dz}{z}}=-\int{dx+c}\]
\[\Rightarrow \log z=-x+c\]
\[\Rightarrow \log |f(x)|=-x+c\]
\[\because f(x)=2,when,x=0\Rightarrow \log 2=c\]
\[\therefore \log |f(x)|=-x+\log 2\]
\[\Rightarrow \log \left| \frac{f(x)}{2} \right|=-x\Rightarrow f(x)=2{{e}^{-x}}\]

\[Solve:{{x}^{4}}\frac{dy}{dx}+{{x}^{3}}y+\cos ec\left( xy \right)=0,given,y=\frac{\pi }{2}when,x=1\]
Solution:
The given equation can be written as
\[{{x}^{3}}\left( xdy+ydx \right)=-\cos ec\left( xy \right)dx\]
\[\Rightarrow \sin \left( xy \right)d\left( xy \right)=-\frac{dx}{{{x}^{3}}}\]
Integrating we get,
\[\int{\sin \left( xy \right)d\left( xy \right)}=-\int{\frac{dx}{{{x}^{3}}}+c}\]
\[\Rightarrow -\cos \left( xy \right)=\frac{1}{2{{x}^{2}}}+c\]
\[Given,y=\frac{\pi }{2}when,x=1\]
\[\therefore 0=\frac{1}{2}+c\Rightarrow c=-\frac{1}{2}\]
Therefore the required particular solution is
\[-\cos \left( xy \right)=\frac{1}{2{{x}^{2}}}-\frac{1}{2}\]
\[\therefore 2{{x}^{2}}\cos \left( xy \right)={{x}^{2}}-1\]

Show that all curves for which the length of the normal is equal to the radius vector are either circles or rectangular hyperbola.
Solution:
Since the length of the normal \[=y\sqrt{1+{{y}_{1}}^{2}}\]
and the radius vector \[=\sqrt{{{x}^{2}}+{{y}^{2}}}\]
\[\therefore y\sqrt{1+{{y}_{1}}^{2}}=\sqrt{{{x}^{2}}+{{y}^{2}}}\]
\[\Rightarrow {{y}^{2}}\left( 1+{{y}_{1}}^{2} \right)={{x}^{2}}+{{y}^{2}}\]
\[\Rightarrow {{y}^{2}}{{y}_{1}}^{2}={{x}^{2}}\]
\[\Rightarrow y{{y}_{1}}=\pm x\]
\[\Rightarrow \frac{dy}{dx}=\pm \frac{x}{y}\]
\[\Rightarrow xdx\pm ydy=0\]
Integrating we get,
\[\int{xdx}\pm \int{ydy}=c\]
\[\therefore \frac{{{x}^{2}}}{2}\pm \frac{{{y}^{2}}}{2}=\frac{{{a}^{2}}}{2},\left[ where,c=\frac{{{a}^{2}}}{2} \right]\]
\[\therefore {{x}^{2}}\pm {{y}^{2}}={{a}^{2}}\]

Show that the equations of the curves for which the Cartesian sub-tangent is constant can be put in the form\[y=c{{e}^{\frac{x}{a}}}\], where c is an arbitrary constant.
Solution:
We know that the Cartesian sub-tangent\[=y\frac{dx}{dy}=a\]
c being a constant
\[\Rightarrow y=a\frac{dy}{dx}\]
\[\Rightarrow \frac{dx}{a}=\frac{dy}{y}\]
Integrating we get,
\[\int{\frac{dx}{a}}=\int{\frac{dy}{y}-\log c}\]
\[\Rightarrow \frac{x}{a}=\log y-\log c\]
\[\therefore \frac{x}{a}=\log \frac{y}{c}\Rightarrow y=c{{e}^{\frac{x}{a}}}\]

Find the equation of the curve whose slope at any point (x, y) is xy and which passes through the point (0, 1).
Solution:
At any point (x, y), slope of the curve y = f(x) is dy/dx
\[Here,\frac{dy}{dx}=xy\]
\[\Rightarrow \frac{dy}{y}=xdx\]
Integrating we get,
\[\int{\frac{dy}{y}}=\int{xdx+c}\]
\[\Rightarrow \log y=\frac{1}{2}{{x}^{2}}+c\]
c being an arbitrary constant.
This is the general equation of the family of curves having slope xy at any point (x, y).
Since the particular curve passes through the point (0, 1), y = 1, when x = 0
\[\therefore \log 1=0+c\Rightarrow c=0\]
Required equation of the curve is
\[\log y=\frac{1}{2}{{x}^{2}}\]
\[\therefore y={{e}^{\frac{{{x}^{2}}}{2}}}\]

Find the eccentricity and the foci of the curve which satisfies the differential equation \[\left( 1+{{y}^{2}} \right)dx-xydy=0\]and passes through the point (1, 0).
Solution:
\[Given,\left( 1+{{y}^{2}} \right)dx-xydy=0\]
\[\Rightarrow \frac{dx}{x}=\frac{ydy}{1+{{y}^{2}}}\]
Integrating we get,
\[\int{\frac{dx}{x}}=\int{\frac{ydy}{1+{{y}^{2}}}+\log c}\]
\[\Rightarrow \log x=\frac{1}{2}\log \left( 1+{{y}^{2}} \right)+\log c\]
\[\Rightarrow \log x=\log \left\{ c\sqrt{1+{{y}^{2}}} \right\}\]
\[\therefore x=c\sqrt{1+{{y}^{2}}}\Rightarrow {{x}^{2}}={{c}^{2}}\left( 1+{{y}^{2}} \right)\]
This is the general equation of the family of curves.
If the curve passes through the point (1, 0), c² = 1
Required equation of the curve is
\[{{x}^{2}}-{{y}^{2}}=1\]
This represents a rectangular hyperbola, here a = b = 1, eccentricity = √2 and the foci are at (±√2, 0).

Show that the curve in which the angle between the tangent and the radius vector at every point is one-half of the vectorial angle is a cardiode.
Solution:
For the curve r = f(θ), if φ be the angle between the tangent and the radius vector at any point, then,
\[\tan \phi =r\frac{d\theta }{dr}\]
\[Here,\phi =\frac{1}{2}\theta \Rightarrow \tan \phi =\tan \frac{\theta }{2}\]
\[\Rightarrow r\frac{d\theta }{dr}=\tan \frac{\theta }{2}\]
\[\Rightarrow \frac{dr}{r}=\cot \frac{\theta }{2}d\theta \]
Integrating we get,
\[\int{\frac{dr}{r}}=\int{\cot \frac{\theta }{2}d\theta +\log c}\]
\[\Rightarrow \log r=2\log \sin \frac{\theta }{2}+\log c\]
\[\Rightarrow \log \frac{r}{c}=\log {{\sin }^{2}}\frac{\theta }{2}\Rightarrow r=c\frac{1}{2}.2{{\sin }^{2}}\frac{\theta }{2}\]
\[\therefore r=\frac{c}{2}\left( 1-\cos \theta  \right)\]
\[\therefore r=a\left( 1-\cos \theta  \right),\left[ where,a=\frac{c}{2} \right]\]
Which represents a cardiode.

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