Question

The general solution of the differential equation $\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}}+x y \frac{d y}{d x}=0$ is (where $C$ is a constant of integration)

• A. $\sqrt{1+y^{2}}+\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left(\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right)+C$
• B. $\sqrt{1+y^{2}}-\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left(\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right)+C$
• C. $\sqrt{1+y^{2}}+\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left(\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right)+C$
• D. $\sqrt{1+y^{2}}-\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left(\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right)+C$

Answer 1

Answer: (A)

$\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}}+x y \frac{d y}{d x}=0$
$\Rightarrow \sqrt{(1+x)^{2}\left(1+y^{2}\right)}+x y \frac{d y}{d x}=0$
$\Rightarrow \sqrt{1+x^{2}} \sqrt{1+y^{2}}=-x y \frac{d y}{d x}$
$\Rightarrow \int \frac{ ydy }{\sqrt{1+ y ^{2}}}=-\int \frac{\sqrt{1+ x ^{2}}}{ x } dx.....(1)$
Now put $1+x^{2}=u^{2}$ and $1+y^{2}=v^{2}$
$2 x d x=2 u d u$ and $2 y d y=2 v d v$
$\Rightarrow xdx = udu$ and $ydy = vdv$
substitude these values in equation (1)
$\int \frac{ vdv }{ v }=-\int \frac{ u ^{2} \cdot du }{ u ^{2}-1}$
$\Rightarrow \int dv =-\int \frac{ u ^{2}-1+1}{ u ^{2}-1} du$
$\Rightarrow v =-\int\left(1+\frac{1}{ u ^{2}-1}\right) du$
$\Rightarrow v=-u-\frac{1}{2} \log _{e}\left|\frac{u-1}{u+1}\right|+c $
$\Rightarrow \sqrt{1+y^{2}}=-\sqrt{1+x^{2}}+\frac{1}{2} \log _{e}\left|\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}-1}}\right|+c $
$\Rightarrow \sqrt{1+y^{2}}+\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left|\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}-1}}\right|+c$