Question

If a curve passing through $\left(1,2\right)$ satisfies the differential equation $y\left(1+xy\right)dx-xdy=0$ , then which of the following is true?

• 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)

The given differential equation is
$y\left(1+xy\right)dx-xdy=0$
$\Rightarrow \left(ydx-xdy\right)+x{y}^{2}dx=0$
$\Rightarrow \frac{ydx-xdy}{y^2}+xdx=0$
$\Rightarrow d\left(\frac{x}{y}\right)+xdx=0$
On integrating both sides, we get
$\int d\left(\frac{x}{y}\right)+\int xdx=0$
$\Rightarrow \frac{x}{y}+\frac{x^2}{2}=C$
If the above curve passes through the point $\left(1,2\right)$ , then
$\frac{1}{2}+\frac{1}{2}=C\Rightarrow C=1$
The given curve is $\frac{x}{y}+\frac{x^2}{2}=1$
$\therefore y=\frac{2x}{2-x^2}$
$\Rightarrow f\left(x\right)=\frac{2x}{2-x^2}$