Question

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

• A. $|x^2-y^2|=|x|$
• B. $|x^2-y^2|=k|x|$, where k is a positive constant.
• C. $|x^2-y^2|=k$, where k is a positive constant.
• D. $|x^2-y^2|=k|x-y|$, k is a positive constant.

Answer 1

The correct option is B $|x^2-y^2|=k|x|$, where k is a positive constant.

$(x^2+y^2)dx-2xydy=0$ is the given equation. Before we start solving it, we must know what type of differential equation it is. The given equation can be re-written as $\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$.
Now $\frac{x^2+y^2}{2xy}$ is a homogeneous function of degree zero. So it is a homogeneous differential equation and hence we will use the method used for homogeneous differential equations.
So put $y=vx$.
So $\frac{dy}{dx}=v+x\frac{dv}{dx}$(Chain rule of differentiation)
$v+x\frac{dv}{dx}=\frac{x^2(1+v^2)}{2x^{2}v}=\frac{1+v^2}{2v}\Rightarrow \frac{xdv}{dx}=\frac{1+v^2}{2v}-v=\frac{1-v^2}{2v}\Rightarrow \frac{2v}{1-v^2}dv=\frac{dx}{x}\cdots \left(1\right)$
Now we have two variables $v$ and $x$. All the terms including $v$ and $x$ can be written separately. So we can use variable separable here onwards.
Integrating (1) we get
$-\log |(1-v^2)|-\log |x|=-\log |c||x(1-v^2)|=|c|=k$ (k > 0)
$\Rightarrow \left|\frac{x(x^2-y^2)}{x^2}\right|=k.\Rightarrow |x^2-y^2|=k|x|$. which is the required solution.