Question

If the curve $y=y\left(x\right)$ satisfies the differential equation $\frac{dy}{dx}-\frac{2xy}{1+x^2}=0$ and $y\left(0\right)=1,$ then $y\left(1\right)$ is equal to

• A. $1$
• B. $\frac{1}{2}$
• C. $2$
• D. $0$

Answer 1

The correct option is C $2$

$\frac{dy}{dx}-\frac{2xy}{1+x^2}=0$
$\Rightarrow \frac{dy}{y}=\frac{2xdx}{1+x^2}\Rightarrow \int \frac{dy}{y}=\int \frac{2xdx}{1+x^2}$

Let $I=\int \frac{2xdx}{1+x^2}$
Assuming $1+x^2=t\Rightarrow 2xdx=dt$
$I=\int \frac{dt}{t}\Rightarrow I=\ln |t|=\ln |1+x^2|$
Hence, the general solution is $\ln |y|=\ln |1+x^2|+\ln |C|$
$\Rightarrow y=C(1+x^2)$
Putting $x=0,y=1$, we get
$C=1$
So, the required curve is,
$y=1+x^2$
$\therefore y\left(1\right)=2$