Question

If $y=f\left(x\right)$ passing through ($1,2$) satisfies the differential equation $y(1+xy)dx-xdy=0$ then

• A. $f\left(x\right)=\frac{2x}{2-x^2}$
• B. $f\left(x\right)=\frac{x+1}{x^2+1}$
• C. $f\left(x\right)=\frac{x-1}{4-x^2}$
• D. $f\left(x\right)=\frac{4x}{1-2x^2}$

Answer 1

Answer: (A)

$f\left(x\right)=\frac{2x}{2-x^2}$

$\Rightarrow \frac{dy}{dx}=\frac{y}{x}+y^2$

$\Rightarrow \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}\frac{1}{x}=1$

Let $-\frac{1}{y}=t\Rightarrow \frac{1}{y^2}\frac{dy}{dx}=\frac{dt}{dx}$

$\Rightarrow \frac{dt}{dx}+t.\frac{1}{x}=1$

Integrating factor =IF =$e^{\frac{1}{x}dx}=e^{lnx}=x$

$\Rightarrow tx=\int x.1dx$

$\Rightarrow \frac{-1}{y}x=\frac{x^2}{2}+c$

Given $y\left(1\right)=2$

$\Rightarrow -\frac{1}{2}=\frac{1}{2}+c$

$\Rightarrow c=-1$

Final solution :

$y=\frac{2x}{2-x^2}$