Problem source

By substituting $y(x)=xv(x)$ in the differential equation 
\[
x^{3}\frac{\mathrm{d}v}{\mathrm{d}x}+x^{2}v=\frac{1+x^{2}v^{2}}{\left(1+x^{2}\right)v},
\]
or otherwise, find the solution $v(x)$ that satisfies $v=1$ when
$x=1$. 

What value does this solution approach when $x$ becomes large?

Solution source

Let $y = xv$ then $y' = v + xv'$ and so $x^2y' = x^2v + x^3v'$ 
Our differential equation is now:

\begin{align*}
&& x^2 y' &= \frac{1+y^2}{(1+x^2)\frac{y}{x}} \\
\Rightarrow && xy' &= \frac{(1+y^2)}{(1+x^2)y} \\
\Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x(1+x^2)} \\
\Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x} - \frac{x}{1+x^2} \\
\Rightarrow && \frac12 \ln(1+y^2) &= \ln x - \frac12 \ln(1+x^2) + C \\
\Rightarrow && \frac12 \ln (1 + y^2) &= \frac12 \ln \l \frac{x^2}{1+x^2}\r + C \\
\Rightarrow && 1+y^2 &= \frac{Dx^2}{1+x^2} \\
\Rightarrow && D &= 4 \quad \quad: (x = 1, v = 1, y = 1) \\
\Rightarrow && 1 + x^2v^2&= \frac{4x^2}{1+x^2}\\
\Rightarrow && v^2 &= \frac{3x^2-1}{x^2(1+x^2)} \\
\Rightarrow && v &= \sqrt{\frac{3x^2-1}{x^2(1+x^2)}} \\
\end{align*}
As $x \to \infty$, $v \to 0$