Question

Solution of differential equation $(x^{2}+y^{2})dx-2xy~dy=0,$ where c is constant, is

• A. $x^{2}+y^{2}=cy$
• B. $x^{2}-y^{2}=cy$
• C. $x^{2}+y^{2}=c~x$
• D. $x^{2}-y^{2}=cx$

Hint:

For homogeneous equations of the form $Mdx + Ndy = 0$, you can also test if the differential is exact. If not, the substitution $y=vx$ is the standard path. Always check if a simple algebraic manipulation like $x^2 - y^2$ appears in the options to guide your final simplification.

Answer 1

The Correct Option is D

This is a homogeneous differential equation because every term in the coefficients of $dx$ and $dy$ is of degree 2.

Step 1: \color{redRewrite in Standard Form
$\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$.
Divide numerator and denominator by $x^2$:
$\frac{dy}{dx} = \frac{1 + (y/x)^2}{2(y/x)}$.

Step 2: \color{redSubstitute $y = vx$
Let $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting into the equation:
$v + x\frac{dv}{dx} = \frac{1 + v^2}{2v}$
$x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v$
$x\frac{dv}{dx} = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}$.

Step 3: \color{redSeparate Variables and Integrate
$\frac{2v}{1 - v^2} dv = \frac{1}{x} dx$
Multiply by -1 on both sides: $\int \frac{-2v}{1 - v^2} dv = \int -\frac{1}{x} dx$
Let $u = 1 - v^2 \implies du = -2v dv$.
$\ln|1 - v^2| = -\ln|x| + \ln|c|$
$\ln|1 - v^2| + \ln|x| = \ln|c|$
$\ln|x(1 - v^2)| = \ln|c| \implies x(1 - v^2) = c$.

Step 4: \color{redBack-substitute and Simplify
Substitute $v = y/x$ back into the result:
$x(1 - \frac{y^2}{x^2}) = c$
$x(\frac{x^2 - y^2}{x^2}) = c$
$\frac{x^2 - y^2}{x} = c \implies x^2 - y^2 = cx$.
This matches Option (4).