Question

Solve the differential equation: $(x^2-2x+2y^2)dx+2xydy=0$

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

Answer 1

The correct option is A $y^2.x^2=\frac{2}{3}{x}^3-\frac{x^4}{4}+c$

$(x^2-2x+{2y}^2)dx+2xydy=0$$\Rightarrow 2y\frac{dy}{dx}+\frac{{2y}^2}{x}=\frac{{-x}^2-2x}{x}$
Put $y^2=v\Rightarrow 2ydy=dv.$
$\therefore \frac{dv}{dx}+\frac{2}{x}v=-x+2$ ...(1)
Here $P=\frac{2}{x}\Rightarrow \int Pdx=\int \frac{2}{x}dx=2\log x$
$\therefore IF=e^{\log x^2}=x^2$
Multiplying (1) throughout by $I.F.,$ we get
$x^2\frac{dv}{dx}+2x=-x^3+{2x}^2$
Integrating both sides
$v.x^3=\int x^2(2-x)dx+c$

$\Rightarrow x^2{y}^2=\frac{2}{3}{x}^3-\frac{x^4}{4}+c$