Question

The solution of the differential equation $\left(x^2-y^2\right)dx+2xydy=0$ is

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

Answer 1

The correct option is D

$x^2+y^2=Cx$

Explanation for the correct option:

The given differential equation: $\left(x^2-y^2\right)dx+2xydy=0$.

$\begin{array}{rcl}& \Rightarrow & 2xydy=-\left(x^2-y^2\right)dx\\ & \Rightarrow & \frac{dy}{dx}=-\frac{x^2-y^2}{2xy}\end{array}$

Let us assume that, $y=ux\_\left(\mathrm{i}\right)$

Differentiate both sides of the equation with respect to $x$.

$\begin{array}{rcl}& \Rightarrow & \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(ux\right)\\ & \Rightarrow & \frac{dy}{dx}=u\frac{dx}{dx}+x\frac{du}{dx}\\ & \Rightarrow & \frac{dy}{dx}=u+x\frac{du}{dx}\_\left(\mathrm{ii}\right)\end{array}$

Consider the equation: $\frac{dy}{dx}=-\frac{x^2-y^2}{2xy}$

From equation $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$.

$\begin{array}{rcl}& \Rightarrow & \left(u+x\frac{du}{dx}\right)=-\frac{x^2-u^2{x}^2}{2x\left(ux\right)}\\ & \Rightarrow & u+x\frac{du}{dx}=-\frac{x^2\left(1-u^2\right)}{2u{x}^2}\\ & \Rightarrow & x\frac{du}{dx}=-\frac{\left(1-u^2\right)}{2u}-u\\ & \Rightarrow & x\frac{du}{dx}=\frac{-\left(1-u^2\right)-2u^2}{2u}\\ & \Rightarrow & x\frac{du}{dx}=\frac{-1-u^2}{2u}\\ & \Rightarrow & \frac{x}{dx}=-\frac{1+u^2}{2udu}\\ & \Rightarrow & \frac{dx}{x}=-\frac{2udu}{1+u^2}\end{array}$

Let us assume that, $\left(1+u^2\right)=t$.

Differentiate both sides of the equation.

$\begin{array}{rcl}& \Rightarrow & d\left(1+u^2\right)=d\left(t\right)\\ & \Rightarrow & 2udu=dt\end{array}$

Now, consider the equation: $\frac{dx}{x}=-\frac{2udu}{1+u^2}$

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

Integrate both sides of the equation.

$\begin{array}{rcl}& \Rightarrow & \int \frac{dx}{x}=\int -\frac{dt}{t}\\ & \Rightarrow & \log \left|x\right|=-\log \left|t\right|+C_1\left[C_1=\mathrm{Integrating}\mathrm{constant}\right]\\ & \Rightarrow & \log \left|xt\right|=C_1\\ & \Rightarrow & \left|xt\right|=e^{C_1}\\ & \Rightarrow & \left|x\left(1+u^2\right)\right|=e^{C_1}\left[\because \left(1+u^2\right)=t\right]\\ & \Rightarrow & \left|x\left(1+{\left(\frac{y}{x}\right)}^2\right)\right|=e^{C_1}\\ & \Rightarrow & x\frac{x^2+y^2}{x^2}=C\left[C=e^{C_1}\right]\\ & \Rightarrow & x^2+y^2=Cx\end{array}$

Therefore, the solution of the differential equation $\left(x^2-y^2\right)dx+2xydy=0$ is $x^2+y^2=Cx$.

Hence, option D is correct.