Question

Solution of differential equation
$\left(x^{2}-2x+2y^{2}\right) \,dx+2xy\,dy=0$ 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 $(x^2 - 2x + 2y^2)\, dx = -2xy \,dy$
$\Rightarrow 2xy \frac{dy}{dx}+2y^{2}=2x-x^{2}$
$\Rightarrow x\left(2y \frac{dy}{dx}\right)+2y^{2}=2x-x^{2}$
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
$vx^{2}=\int x^{2}\left(2-x\right)dx$
$vx^{2}=\frac{2x^{3}}{3}-\frac{x^{4}}{4}+c$
$\Rightarrow v=\frac{2x}{3}-\frac{1}{4}x^{2}+\frac{c}{x^{2}}$
$\therefore y^{2}=\frac{2x}{3}-\frac{1}{4}x^{2}+\frac{c}{x^{2}}$ which is required solution.