Question

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

• A. $\begin{array}{l}\sqrt{1+y^2}+\sqrt{1+x^2}=\frac{1}{2}lo{g}_e\left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right)+c\end{array}$
• B. $\begin{array}{l}\sqrt{1+y^2}-\sqrt{1+x^2}=\frac{1}{2}lo{g}_e\left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right)+c\end{array}$
• C. $\begin{array}{l}\sqrt{1+y^2}+\sqrt{1+x^2}=\frac{1}{2}lo{g}_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)+c\end{array}$
• D. $\begin{array}{l}\sqrt{1+y^2}-\sqrt{1+x^2}=\frac{1}{2}lo{g}_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)+c\end{array}$

Answer 1

The correct option is C

$\begin{array}{l}\sqrt{1+y^2}+\sqrt{1+x^2}=\frac{1}{2}lo{g}_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)+c\end{array}$

Step 1: Doing Variable Separation

The given equation is

$\sqrt{1+x^2+y^2+x^2{y}^2}+xy\frac{dy}{dx}=0$

$\Rightarrow$$\sqrt{1\left(1+x^2\right)+y^2\left(1+x^2\right)}+xy\frac{dy}{dx}=0$

$\Rightarrow$ $\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+xy\frac{dy}{dx}=0$

$\Rightarrow$ $\frac{\sqrt{1+x^2}dx}{x}=-\frac{ydy}{\sqrt{1+y^2}}$

Step 2: Integrating the Expression

Upon integrating both sides of the equation we get,

$\int \frac{\sqrt{1+x^2}dx}{x}=-\int \frac{ydy}{\sqrt{1+y^2}}$ $\left(1\right)$

Let $1+x^2=t^2$

$\therefore 2xdx=2tdt$

$\Rightarrow$ $dx=\frac{t}{x}dt$

Similarly, let

$1+y^2=z^2$

$\therefore 2ydy=2zdz$

$\Rightarrow$ $dy=\frac{z}{y}dz$

Upon substituting these in equation $\left(1\right)$ we get,

$\int \frac{t.tdt}{t^2-1}=-\int \frac{zdz}{z}$

$\Rightarrow$ $\int \frac{\left(t^2-1+1\right)dt}{t^2-1}=-z+c$

$\Rightarrow$ $\int dt+\int \frac{dt}{t^2-1}=-z+c$

$\Rightarrow$ $t+\frac{1}{2}{\log }_e\left(t^2-1\right)=-z+c$

$\Rightarrow$ $t+\frac{1}{2}{\log }_e\frac{\left(t-1\right)}{\left(t+1\right)}=-z+c$

$\Rightarrow$$\sqrt{1+x^2}+\frac{1}{2}{\log }_e\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}=-\sqrt{1+y^2}+c$

$\Rightarrow$ $\begin{array}{l}\sqrt{1+y^2}+\sqrt{1+x^2}=\frac{1}{2}lo{g}_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)+c\end{array}$

Hence, option $C$ is the correct answer.