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

Answer 1

Answer: (C)

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