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Q.
The solution of the differential equation $x d y-y d x=\sqrt{x^{2}+y^{2}} d x$, given that $y=1$ when $x=\sqrt{3}$, is
TS EAMCET 2020
Solution:
Given, differential equation
$x d y-y d x=\sqrt{x^{2}+y^{2}} d x \ldots (i)$
Putting, $y=v x$ and differentiating w.r.t. $x$, we get
$d y=v d x +x d v \ldots \text { (ii) }$
From Eq. (ii),
$x(v d x+x d v)-v x d x=\sqrt{x^{2}+(v x)^{2}} d x$
$x v d x+x^{2} d v-v x d x=x \sqrt{1+v^{2}} d x$
$x^{2} d v=x \sqrt{1+v^{2}} d x$
$\Rightarrow \frac{d v}{\sqrt{1+v^{2}}}=\frac{d x}{x}$
Integrating on both sides, we get
$\ln \left(v+\sqrt{1+v^{2}}\right)=\ln x+ C$
$\left\{\because \int \frac{d x}{\sqrt{1+x^{2}}}=\ln \left(x+\sqrt{1+x^{2}}\right)+C\right\}$
Putting $v=\frac{y}{x}$, we get
$\ln \left(\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^{2}}\right)=\ln x +C$
$\ln \left(\frac{y}{x}+\frac{1}{x} \sqrt{x^{2}+y^{2}}\right)=\ln x +C$
$\ln \left\{\frac{1}{x}\left(y+\sqrt{x^{2}+y^{2}}\right)\right\}-\ln x=C$
$\ln \left\{\frac{y+\sqrt{x^{2}+y^{2}}}{x^{2}}\right\}=C \ldots (iii)$
$\left[\because \ln a-\ln b=\ln \frac{a}{b}\right]$
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
$\left(x^{2}-y\right)^{2}=x^{2}+y^{2}$