Question

The solution of the differential equation $xdy-ydx=\sqrt{x^2+y^2}dx$, given that $y=1$ when $x=\sqrt{3}$, is

• A. ${\left(x^2-y^2\right)}^2=x^2+y^2$
• B. ${\left(x^2-y^2\right)}^2=x^2+y^2$
• C. ${\left(x^2+y\right)}^2=x^2-y^2$
• D. $x^2-y={\left(x+y^2\right)}^2$

Answer 1

Answer: (B)

Given, differential equation
$xdy-ydx=\sqrt{x^2+y^2}dx\dots (i)$
Putting, $y=vx$ and differentiating w.r.t. $x$, we get
$dy=vdx+xdv\dots \text{ (ii) }$
From Eq. (ii),
$x(vdx+xdv)-vxdx=\sqrt{x^2+(vx{)}^2}dx$
$xvdx+x^{2}dv-vxdx=x\sqrt{1+v^2}dx$
$x^{2}dv=x\sqrt{1+v^2}dx$
$\Rightarrow \frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x}$
Integrating on both sides, we get
$\ln \left(v+\sqrt{1+v^2}\right)=\ln x+C$
$\left\{\because \int \frac{dx}{\sqrt{1+x^2}}=\ln \left(x+\sqrt{1+x^2}\right)+C\right\}$
Putting $v=\frac{y}{x}$, we get
$\ln \left(\frac{y}{x}+\sqrt{1+{\left(\frac{y}{x}\right)}^2}\right)=\ln x+C$
$\ln \left(\frac{y}{x}+\frac{1}{x}\sqrt{x^2+y^2}\right)=\ln x+C$
$\ln \left\{\frac{1}{x}\left(y+\sqrt{x^2+y^2}\right)\right\}-\ln x=C$
$\ln \left\{\frac{y+\sqrt{x^2+y^2}}{x^2}\right\}=C\dots (iii)$
$\left[\because \ln a-\ln b=\ln \frac{a}{b}\right]$
Given that, $y=1,x=\sqrt{3}$
$\therefore \ln \left\{\frac{1+\sqrt{(\sqrt{3}{)}^2+1^2}}{(\sqrt{3}{)}^2}\right\}=C$ $\Rightarrow \ln \left\{\frac{1+2}{3}\right\}=C$
$\ln l=C\text{ or }C=0[\because \ln 1=0]$
Putting the value of $C$ in Eq. (iii), we get
$\ln \left\{\frac{y+\sqrt{x^2+y^2}}{x^2}\right\}=0$
or $\frac{y+\sqrt{x^2+y^2}}{x^2}=e^0=1$
$y+\sqrt{x^2+y^2}=x^2$
or $x^2-y=\sqrt{x^2+y^2}$
Squaring on both sides, we get
${\left(x^2-y\right)}^2=x^2+y^2$