Question

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

• A. $y^2=2x-\frac{1}{4}{x}^2+\frac{c}{x^2}$
• B. $y^2=\frac{2}{3}x-x^2+\frac{c}{x^2}$
• C. $y^2=\frac{2}{3}x-\frac{x^2}{4}+\frac{c}{x^2}$
• D. none of these

Answer 1

Answer: (C)

As $(x^2-2x+2y^2)dx=-2xydy$
$\Rightarrow 2xy\frac{dy}{dx}+2y^2=2x-x^2$
$\Rightarrow x\left(2y\frac{dy}{dx}\right)+2y^2=2x-x^2$
By putting $y^2=v$
$\Rightarrow 2y\cdot \frac{dy}{dx}=\frac{dv}{dx}$
$\Rightarrow x\frac{dv}{dx}+2v=2x-x^2$
$\Rightarrow \frac{dv}{dx}+v\left(\frac{2}{x}\right)=2-x$
$I.F.=e^{\int \frac{2}{x}dx}=x^2$
Now, required solution is
$v{x}^2=\int x^2\left(2-x\right)dx$
$v{x}^2=\frac{2x^3}{3}-\frac{x^4}{4}+c$
$\Rightarrow v=\frac{2x}{3}-\frac{1}{4}{x}^2+\frac{c}{x^2}$
$\therefore y^2=\frac{2x}{3}-\frac{1}{4}{x}^2+\frac{c}{x^2}$ which is required solution.