Question

The solution of differential equation $\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$ is

• A. $\sqrt{x^2+y^2}=a\sin ({\tan }^{-1}\frac{y}{x}+c)$
• B. $\sqrt{x^2+y^2}=a\sin (k-{\cot }^{-1}\frac{y}{x})$
• C. $x=y\cot (\sin \frac{\sqrt{x^2+y^2}}{a}+c)$, if $xy>0$
• D. $y=x\tan (\sin \frac{\sqrt{x^2+y^2}}{a}+c)$, if $xy>0$

Answer 1

The correct option is A $\sqrt{x^2+y^2}=a\sin ({\tan }^{-1}\frac{y}{x}+c)$

Given,

$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$

Let x=$rcos\theta$

$dx=cos\theta dr-rsin\theta d\theta$

$dy=rsin\theta d\theta +cos\theta dr$

Taking L.H.S. of given equation,

$\frac{rcos\theta (cos\theta dr-rsin\theta d\theta )-rsin\theta (rsin\theta d\theta +cos\theta dr)}{rcos\theta (rsin\theta d\theta +cos\theta dr)-rsin\theta (cos\theta dr-rsin\theta d\theta }$

$\to \frac{dr(rco{s}^2\theta +rsi{n}^2\theta )+d\theta (r^{2}sin\theta cos\theta -r^{2}sin\theta cos\theta )}{dr(rsin\theta cos\theta -rsin\theta cos\theta )+d\theta (r^{2}co{s}^2\theta +r^{2}si{n}^2\theta )}$

$\to \frac{rdr}{r^{2}d\theta }$

Taking R.H.S. of ggiven equation,

$\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$

$\to \sqrt{\frac{a^2-r^2}{r^2}}$

Since L.H.S.=R.H.S.

$\frac{dr}{rd\theta }=\frac{\sqrt{a^2-r^2}}{r}$

$\frac{dr}{\sqrt{a^2-r^2}}=d\theta$

Integrating Both the sides

$\int \frac{dr}{\sqrt{a^2-r^2}}=\int d\theta$

$si{n}^{-1}\left(\frac{r}{a}\right)=\theta +c$

$\frac{r}{a}=sin(\theta +c)$

$\sqrt{x^2+y^2}=asin\left(ta{n}^{-1}\right(\frac{y}{x})+c)$