Question

The equation of the curve satisfying the differential equation $y_{2}\left(x^{2}+1\right)=2 x y_{1}$ passing through the point $(0,1)$ and having slope of tangent at $x=0$ as $3$ is

• A. $y=x^{2}+3 x+2$
• B. $y^{2}=x^{2}+3 x+1$
• C. $y=x^{3}+3 x+1$
• D. None of these

Answer 1

Answer: (C)

The given differential equation is
$y_{2}\left(x^{2}+1\right)=2 x y_{1} \Rightarrow \frac{y_{2}}{y_{1}}=\frac{2 x}{x^{2}+1}$
Integrating both sides, we get
$\log y_{1}=\log \left(x^{2}+1\right)+\log C \Rightarrow y_{1}=C\left(x^{2}+1\right)$
It is given that $y_{1}=3$ at $x=0$
Putting $x=0, y_{1}=3$ in (i), we get $3=C$
Substituting the value of $C$ in (i), we obtain
$y_{1}=3\left(x^{2}+1\right)$
Integrating both sides with respect to $x,$ we get
$y=x^{3}+3 x+C_{2}$
This passes through the point $(0,1)$ . Therefore, $1=C_{2}$
Hence, the required equation of the curve is $y=x^{3}+3 x+1$