Question

Let $(x^2+1)\frac{d^{2}y}{d{x}^2}=2x\frac{dy}{dx},$ where $y^{\prime }\left(0\right)=3$ & $y\left(0\right)=1$, then $y\left(1\right)$ is equal to:

Answer 1

$\frac{y^{\prime \prime }}{y^{\prime }}=\frac{2x}{x^2+1}$
On integrating, we get $\ln y^{\prime }=\ln (x^2+1)+\ln c$
$\Rightarrow y^{\prime }=c(x^2+1)$
$y^{\prime }\left(0\right)=3$

$\Rightarrow c=3$
$y^{\prime }=3(x^2+1)$
$\Rightarrow y=x^3+3x+1$

$\Rightarrow y\left(1\right)=5$