Solve the differential equation. $y^\prime = y^2+\frac{1}{x^4}$, $y=\frac{1}{x^2}\text{ctg}(\frac{1}{x}+c) - \frac{1}{x}$ and $y(+\infty)=0$ Solve the differential equation.
$$y^\prime = y^2+\frac{1}{x^4}.$$
I am given this as a solution.Need to choose one which satisfies given condition.($y(+\infty)=0$)
$$y=\frac{1}{x^2}\cot(\frac{1}{x}+c) - \frac{1}{x}$$ and $$y(+\infty)=0.$$
We need to find $c$ and show that it is solution for differential equation.
So
$$\lim_{x\to +\infty}\frac{1}{x^2}\cot(\frac{1}{x}+c) - \frac{1}{x} =0
\\
\lim_{x\to+\infty}\frac{1}{x^2}\frac{\cos(\frac{1}{x}+c)}{\sin(\frac{1}{x}+c)} =0$$
From here I don't know how to find $c$. Will be glad if you can help me.
 A: Let's make the variable change
\begin{equation}
t = \frac{1}{x}
\end{equation}
Then
\begin{equation}
\lim_{x\rightarrow +\infty}\left(\frac{1}{x^{2}}\cdot\cot\left(\frac{1}{x} + c\right) - \frac{1}{x}\right) = \lim_{t\rightarrow 0}\left(t^{2}\cot\left(t + c\right) - t\right) = 0
\end{equation}
Therefore any value of $c$ will satisfies this equation.
Next for the left part of equation
\begin{equation}
y’ = - \frac{2\cot\left(\frac{1}{x} + c\right)}{x^{3}} + \frac{1}{x^{4}\sin^{2}\left(\frac{1}{x} + c\right)} + \frac{1}{x^{2}}
\end{equation}
For the right part
\begin{multline}
y^2 + \frac{1}{x^{4}} = \frac{\cot^{2}\left(\frac{1}{x} + c\right)}{x^{4}} - \frac{2\cot\left(\frac{1}{x} + c\right)}{x^{3}} + \frac{1}{x^{4}} + \frac{1}{x^{4}} = \\ = \frac{\cot^{2}\left(\frac{1}{x} + c\right) + 1}{x^{4}} - \frac{2\cot\left(\frac{1}{x} + c\right)}{x^{3}} + \frac{1}{x^{4}} = \\ =\frac{\cos^{2}\left(\frac{1}{x} + c\right) + \sin ^{2}\left(\frac{1}{x} + c\right)} {x^{4}\sin^{2}\left(\frac{1}{x} + c\right)} - \frac{2\cot\left(\frac{1}{x} + c\right)}{x^{3}} + \frac{1}{x^{4}} = \\ = - \frac{2\cot\left(\frac{1}{x} + c\right)}{x^{3}} + \frac{1}{x^{4}\sin^{2}\left(\frac{1}{x} + c\right)} + \frac{1}{x^{2}}
\end{multline}
Therefore
\begin{equation}
y’ = y^{2} + \frac{1}{x^{4}}
\end{equation}
which means that the given $y$ is a solution for this equation.
