Question

If $2xydy=\left(x^2+y^2+1\right)dx,y(1)=0$ and $y\left(x_0\right)=\sqrt{3}$, then $x_0$ can be

• A. 2
• B. - 2
• C. 3
• D. - 3

Answer 1

Answer: (A), (B)

$2xy\frac{dy}{dx}=x^2+y^2+1$
Put $y^2=t$
$x\frac{dt}{dx}=x^2+t+1$
$\frac{dt}{dx}-\frac{t}{x}=\frac{x^2+1}{x}$
$\text{ I.F. }=\frac{1}{x}$
Hence $\frac{t}{x}=\int \frac{x^2+1}{x^2}dx=x-\frac{1}{x}+C$
$\therefore y^2=x^2-1,C=0$ as $y(1)=0$
Now, $y\left(x_0\right)=\sqrt{3}\Rightarrow 3=x_0^2-1\Rightarrow x_0^2=4$
$\therefore x_0=\pm 2$