# Find general solution of $2x\cos^2(y)+(x^2+1)\sin(y)y'=0$

Find the general solution of $$2x\cos^2(y)+(x^2+1)\sin(y)y'=0.$$

First, I divided by $$\sin(y)$$: $$2x\frac{\cos^2(y)}{\sin(y)}+(x^2+1)y'=0\implies(x^2+1)y'=-2x\frac{\cos^2(y)}{\sin(y)},$$ so $$\frac{\sin(y)}{\cos^2(y)}y'=-\frac{2x}{x^2+1}\implies\int\frac{\sin(y)}{\cos^2(y)}\,\mathrm{d}y=\int-\frac{2x}{x^2+1}\,\mathrm{d}x.$$ For LHS: pick $$u=\cos(y)$$, so $$\mathrm{d}u=-\sin(y)\,\mathrm{d}y$$, then the integral becomes $$-\int\frac{1}{u^2}\,\mathrm{d}u=\frac{1}{u}+C_1=\frac{1}{\cos(y)}+C_1=\sec(y)+C_1.$$ For RHS: pick $$u=x^2+1$$, then $$\mathrm{d}u=2x\,\mathrm{d}x$$, so the integral becomes $$-\int\frac{1}{u}\,\mathrm{d}u=-\ln\lvert u\rvert+C_2=-\ln\lvert x^2+1\rvert+C_2\underbrace{=}_{x^2+1>0}-\ln(x^2+1)+C_2.$$ Finally, $$\sec(y)+C_1=-\ln(x^2+1)+C_2\implies\boxed{y=\sec^{-1}(-\ln(x^2+1)+C)}.$$ However, WA gives two solutions:

https://www.wolframalpha.com/input/?i=2x(cos(y))%5E2%2B(x%5E2%2B1)sin(y)y%27%3D0

One of them is my solution, and the other is $$y=-\sec^{-1}(-\ln(x^2+1)+C)$$, which is also true because if we plug it into the differential equation we end up with $$0$$, and $$0=0$$ is true.

What did I miss?

The secant function is even (and not injective). This means that $$\sec(y)=f(x)$$ has multiple solutions $$y$$ for any particular value of $$x$$. Two of these solutions are $$y=\pm \sec^{-1}(f(x))$$
But there are even more solutions than that. Since $$\sec(x)$$ is periodic with period $$2\pi$$, we also have solutions like $$y=2\pi k\pm\sec^{-1}(f(x))$$ where $$k\in\mathbb Z$$.
• Wow, you are right! Thanks! Since we have divided by $\sin(y)$ we cannot expect the values $k\pi$ ($k\in\Bbb{Z}$) for $y$, so the general solution is $y=\pm\sec^{-1}(-\ln(x^2+1)+C)$ but not $y=2\pi k\pm\sec^{-1}(-\ln(x^2+1)+C)$, right? – manooooh Jun 2 at 21:38
• @manooooh Actually, $2\pi k \pm \sec^{-1}(-\ln(x^2+1)+C)$ solves the differential equation for any value of $k$. – Franklin Pezzuti Dyer Jun 2 at 21:40