Question

The solution of the differential equation $2x^{3}ydy+\left(1-y^2\right)\left(x^2{y}^2+y^2-1\right)dx=0$ is
(Note: Where $C$ is a constant)

• A. $x^2{y}^2=(Cx+1)\left(1-y^2\right)$
• B. $x^2{y}^2=(Cx+1)\left(1+y^2\right)$
• C. $x^2{y}^2=(Cx-1)\left(1-y^2\right)$
• D. none of these

Answer 1

Answer: (C)

We have,
$2x^{3}ydy+\left(1-y^2\right)\left(x^2{y}^2+y^2-1\right)dx=0$
$\Rightarrow \left\{\frac{2x^{3}y}{{\left(1-y^2\right)}^2}\right\}dy+\left\{\frac{\left(1-y^2\right)\left(x^2{y}^2+y^2-1\right)}{{\left(1-y^2\right)}^2}\right\}dx=0$
$\Rightarrow \left\{\frac{2x^{3}y}{{\left(1-y^2\right)}^2}\right\}dy+\left\{\frac{\left(x^2{y}^2+y^2-1\right)}{\left(1-y^2\right)}\right\}dx=0$
$\Rightarrow \left\{\frac{2x^{3}y}{{\left(1-y^2\right)}^2}\right\}dy+\left\{\frac{\left(x^2{y}^2-\left(1-y^2\right)\right)}{\left(1-y^2\right)}\right\}dx=0$
$\Rightarrow \left\{\frac{2x^{3}y}{{\left(1-y^2\right)}^2}\right\}dy+\left\{\frac{x^2{y}^2}{\left(1-y^2\right)}-1\right\}dx=0$
$\Rightarrow \frac{2y}{{\left(1-y^2\right)}^2}\frac{dy}{dx}=\frac{1}{x^3}-\frac{y^2}{x\left(1-y^2\right)}$
$\Rightarrow \frac{2y}{{\left(1-y^2\right)}^2}\frac{dy}{dx}+\frac{y^2}{x\left(1-y^2\right)}=\frac{1}{x^3}$
Put $\frac{y^2}{1-y^2}=u\Rightarrow \frac{2y}{{\left(1-y^2\right)}^2}\cdot \frac{dy}{dx}=\frac{du}{dx}$
$\therefore \frac{du}{dx}+\frac{u}{x}=\frac{1}{x^3}$
$\text{I.F.}=e^{\int \frac{dx}{x}}=e^{lnx}=x$
Solution is
$xu=\int x\cdot \frac{1}{x^3}dx+C$
$\Rightarrow xu=\int \frac{1}{x^2}dx+C$
$\Rightarrow xu=-\frac{1}{x}+C$
$\Rightarrow x\left(\frac{y^2}{1-y^2}\right)=-\frac{1}{x}+C$
$\Rightarrow \left(\frac{y^2}{1-y^2}\right)=Cx-1$
$\Rightarrow y^2=\left(Cx-1\right)\left(1-y^2\right)$