3.3.6.7 Example 7 $x^3{y}^{\prime }-\sin y=1$

Solve

$$\begin{array}{rl}{x}^3{y}^{\prime }-\sin y& =1\\ y\left(\mathrm{\infty }\right)& =5\pi \end{array}$$

Writing the ode as

$$y^{\prime }=\frac{1+\sin y}{x^3}x\ne 0$$

Shows it is separable

$$\frac{dy}{1+\sin y}=\frac{dx}{x^3}$$

Integrating gives

$$\begin{array}{rl}\int \frac{dy}{1+\sin y}& =\int \frac{dx}{x^3}\\ \frac{-2}{\tan \left(\frac{y}{2}\right)+1}& =-\frac{1}{2x^2}+c\\ \frac{2}{\tan \left(\frac{y}{2}\right)+1}& =\frac{1}{2x^2}-c\\ \frac{2}{\tan \left(\frac{y}{2}\right)+1}& =\frac{1-2x^{2}c}{2x^2}\\ \tan \left(\frac{y}{2}\right)+1& =\frac{4x^2}{1-2x^{2}c}\\ \tan \left(\frac{y}{2}\right)& =\frac{4x^2}{1-2x^{2}c}-1\\ \tan \left(\frac{y}{2}\right)& =\frac{4x^2-(1-2x^{2}c)}{1-2x^{2}c}\\ \tan \left(\frac{y}{2}\right)& =\frac{4x^2-1+2x^{2}c}{1-2x^{2}c}\end{array}$$

Hence

$$\begin{array}{rl}\frac{y}{2}& =\arctan \left(\frac{4x^2-1+2x^{2}c}{1-2x^{2}c}\right)+\pi nn\in \mathbb{Z}\\ \text{(1)}& y& =2(\arctan \left(\frac{4x^2-1+2x^{2}c}{1-2x^{2}c}\right)+\pi n)\end{array}$$

Applying IC gives, and taking limit $\underset{x\to \mathrm{\infty }}{lim}\left(\frac{4x^2-1+2x^{2}c}{1-2x^{2}c}\right)=-\frac{4+2c}{2c}$ assuming $c\ne 0$ then (1) above becomes

$$\begin{array}{rl}5\pi & =2(\arctan (-\frac{4+2c}{2c})+\pi n)\\ & =2\arctan (-\frac{4+2c}{2c})+2\pi n\\ \frac{5\pi -2\pi n}{2}& =\arctan (-\frac{4+2c}{2c})\\ \frac{2\pi n-5\pi }{2}& =\arctan \left(\frac{4+2c}{2c}\right)\end{array}$$

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 $\underset{x\to \mathrm{\infty }}{lim}\left(\frac{4x^2-1+2x^{2}c}{1-2x^{2}c}\right)$ assuming $c=0$ gives $\mathrm{\infty }$. Hence (1) becomes

$$\begin{array}{rl}5\pi & =2(\arctan \left(\mathrm{\infty }\right)+\pi n)\\ 5\pi & =2\left(\frac{\pi }{2}\right)+2\pi n\\ 5\pi & =\pi +2\pi n\\ 5\pi -\pi & =2\pi n\\ n& =2\end{array}$$

Hence (1) becomes (using $c=0,n=2$)

$$\begin{array}{rl}y& =2(\arctan (4x^2-1)+2\pi )\\ & =2\arctan (4x^2-1)+4\pi \end{array}$$

This solution satisfies the ode now and the IC.