Solve\begin {align*} x^{3}y^{\prime }-\sin y & =1\\ y\left ( \infty \right ) & =5\pi \end {align*}
Writing the ode as \[ y^{\prime }=\frac {1+\sin y}{x^{3}}\qquad x\neq 0 \] Shows it is separable\[ \frac {dy}{1+\sin y}=\frac {dx}{x^{3}}\] Integrating gives\begin {align*} \int \frac {dy}{1+\sin y} & =\int \frac {dx}{x^{3}}\\ \frac {-2}{\tan \left ( \frac {y}{2}\right ) +1} & =-\frac {1}{2x^{2}}+c\\ \frac {2}{\tan \left ( \frac {y}{2}\right ) +1} & =\frac {1}{2x^{2}}-c\\ \frac {2}{\tan \left ( \frac {y}{2}\right ) +1} & =\frac {1-2x^{2}c}{2x^{2}}\\ \tan \left ( \frac {y}{2}\right ) +1 & =\frac {4x^{2}}{1-2x^{2}c}\\ \tan \left ( \frac {y}{2}\right ) & =\frac {4x^{2}}{1-2x^{2}c}-1\\ \tan \left ( \frac {y}{2}\right ) & =\frac {4x^{2}-\left ( 1-2x^{2}c\right ) }{1-2x^{2}c}\\ \tan \left ( \frac {y}{2}\right ) & =\frac {4x^{2}-1+2x^{2}c}{1-2x^{2}c} \end {align*}
Hence\begin {align} \frac {y}{2} & =\arctan \left ( \frac {4x^{2}-1+2x^{2}c}{1-2x^{2}c}\right ) +\pi n\qquad n\in \mathbb {Z}\nonumber \\ y & =2\left ( \arctan \left ( \frac {4x^{2}-1+2x^{2}c}{1-2x^{2}c}\right ) +\pi n\right ) \tag {1} \end {align}
Applying IC gives, and taking limit \(\lim _{x\rightarrow \infty }\left ( \frac {4x^{2}-1+2x^{2}c}{1-2x^{2}c}\right ) =-\frac {4+2c}{2c}\) assuming \(c\neq 0\) then (1) above becomes\begin {align*} 5\pi & =2\left ( \arctan \left ( -\frac {4+2c}{2c}\right ) +\pi n\right ) \\ & =2\arctan \left ( -\frac {4+2c}{2c}\right ) +2\pi n\\ \frac {5\pi -2\pi n}{2} & =\arctan \left ( -\frac {4+2c}{2c}\right ) \\ \frac {2\pi n-5\pi }{2} & =\arctan \left ( \frac {4+2c}{2c}\right ) \end {align*}
The range of \(\arctan \) is \(-\frac {\pi }{2}\) to \(\frac {\pi }{2}\). Hence we need \(\frac {2\pi n-5\pi }{2}\) to be in this range. This means \(2\pi n-5\pi \) should be between \(-\pi \cdots \pi \) but not including the edge points. Value of \(n\) which allows this is \(n=-\frac {5}{2}\). (but \(n\) should be an integer. There is no integer solution.) Hence this leads to no solution.
Now we go back to (1) and take the limit assuming \(c=0\).
Applying IC gives, and taking limit \(\lim _{x\rightarrow \infty }\left ( \frac {4x^{2}-1+2x^{2}c}{1-2x^{2}c}\right ) \) assuming \(c=0\) gives \(\infty \). Hence (1) becomes
\begin {align*} 5\pi & =2\left ( \arctan \left ( \infty \right ) +\pi n\right ) \\ 5\pi & =2\left ( \frac {\pi }{2}\right ) +2\pi n\\ 5\pi & =\pi +2\pi n\\ 5\pi -\pi & =2\pi n\\ n & =2 \end {align*}
Hence (1) becomes (using \(c=0,n=2\))\begin {align*} y & =2\left ( \arctan \left ( 4x^{2}-1\right ) +2\pi \right ) \\ & =2\arctan \left ( 4x^{2}-1\right ) +4\pi \end {align*}
This solution satisfies the ode now and the IC.