Question

The solution of the differential equation $\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{1-x^2-y^2}{x^2+y^2}}$ is:

• A. $\sqrt{x^2+y^2}=sin\{ta{n}^{-1}\left(\frac{y}{x}\right)+c\}$
• B. $\sqrt{x^2+y^2}=cos\left\{ta{n}^{-1}(\frac{y}{x}+c)\right\}$
• C. $\sqrt{x^2+y^2}=tan\left\{si{n}^{-1}(\frac{y}{x}+c)\right\}$
• D. $y=xtan(c+si{n}^{-1}\sqrt{x^2+y^2})$

Answer 1

Answer: (A), (D)

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

$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{1-x^2-y^2}{x^2{+y}^2}}$

The above differential equation can be re-written as,

$\frac{xdx+ydy}{\sqrt{1-{(x}^2+y^2)}}=\frac{xdy-ydx}{\sqrt{x^2+y^2}}$.................(1)

$\because d\left({\tan }^{-1}\left(\frac{y}{x}\right)\right)=\frac{xdy-ydx}{x^2+y^2}$ and $d(x^2+y^2)=2(xdx+ydy)$ .................(2)

multiplying both sides in the denominator of equation (1) with $\sqrt{(x^2+y^2)}$

we get, $\frac{xdx+ydy}{\sqrt{x^2+y^2}\sqrt{1-{(x}^2+y^2)}}=\frac{xdy-ydx}{x^2+y^2}$ ...........................(3)

From (2) and (3), we have

$\frac{\frac{1}{2}d(x^2+y^2)}{\sqrt{x^2+y^2}\sqrt{1-{(x}^2+y^2)}}=d\left({\tan }^{-1}\left(\frac{y}{x}\right)\right)$

Now, put $x^2+y^2=t^2$ in the L.H.S. and we get,

$\frac{\frac{1}{2}d\left(t^2\right)}{t\sqrt{(1-t^2)}}=d\left({\tan }^{-1}\left(\frac{y}{x}\right)\right)$

$\Rightarrow$ $\frac{tdt}{t\sqrt{(1-t^2)}}=d\left({\tan }^{-1}\left(\frac{y}{x}\right)\right)$

$\Rightarrow$ $\frac{dt}{\sqrt{(1-t^2)}}=d\left({\tan }^{-1}\left(\frac{y}{x}\right)\right)$

Now integrating both the sides i.e.

$\int \frac{dt}{\sqrt{(1-t^2)}}=\int d\left({\tan }^{-1}\left(\frac{y}{x}\right)\right)$

$\Rightarrow$ ${\sin }^{-1}t={\tan }^{-1}\left(\frac{y}{x}\right)+C$ (where C is a constant)

Now, put the value of $t$, we get

$C+{\sin }^{-1}\left(\sqrt{x^2+y^2}\right)={\tan }^{-1}\left(\frac{y}{x}\right)$........(4)

$\Rightarrow$ $y=x\tan (C+{\sin }^{-1}\left(\sqrt{x^2+y^2}\right))$

Hence, option (D) is correct.

Again re-writing equation (4),

${\sin }^{-1}\left(\sqrt{x^2+y^2}\right)={\tan }^{-1}\left(\frac{y}{x}\right)+C$

$\Rightarrow$ $\sqrt{(x^2+y^2)}=\sin ({\tan }^{-1}\left(\frac{y}{x}\right)+C)$

Hence, option (A) is also correct.