Question

If $y=\left(x\right)$ 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. $f\left(x\right)=\frac{4x}{x^2+1}$

Answer 1

Answer: (A)

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

Given, $y(1+xy)dx-xdy=0$

$\Rightarrow ydx-xdy=--x{y}^{2}dx$

$\Rightarrow \frac{ydx-xdy}{y^2}=-xdx$

$\Rightarrow d\left(\frac{x}{y}\right)=-d\left(\frac{x^2}{2}\right)$

Integrating both sides, we have

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

Above curve pass through $(1,2)$

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

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

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

Hence, option A is correct.