# Can't figure this out: $dy/dx = \cos(y)\sin(x)$

Problem: Find the solution and domain of validity for the following differential equation: $$\dfrac{dy}{dx} = \cos(y) \cdot \sin(x)$$

What I tried: I noticed that this is a separable equation, so I wrote it in the form: $$\dfrac{1}{\cos(y)} \cdot dy = \sin(x) \cdot dx$$

Next, I found out that if $$\cos(y) = 0$$ then $$\dfrac{1}{\cos(y)}$$ is undefined. So I separate into two following cases;

Case1: $$\cos(y) \neq 0$$;

Then $$\int\dfrac{dy}{\cos(y)} = \int\sin(x)dx$$. By solving these integrals I get that $$\tan(y) \cdot \sec(y)+ c_0 = -\cos(x)$$ which implies that

$$\arccos(-\tan(y) \cdot \sec(y)+ c) = x$$, for $$c=-c_0$$ which is an arbitrary constant.

I wonder if this is a general solution.

Case2: $$\cos(y)=0$$;

This implies that $$y\in \{ \mathbb{R} -\dfrac{(2k+1)\pi}2:k\in \mathbb{Z}\}$$

Pick an arbitrary $$y_0\in \{ \mathbb{R} -\dfrac{(2k+1)\pi}2:k\in \mathbb{Z}\}$$, substituting this into the differential equation gives that;

$$\dfrac {dy_0}{dx} = \cos(y_0)\cdot \sin(x)$$. I am pretty sure that $$\cos(y_0) = 0$$. But I do not know how can I show that $$\dfrac {dy_0}{dx} = 0$$ to find out if there is a singular/lost solution or not.

Moreover, I have been taught that $$dx$$ means derivative, for instance $$x^2\cdot dx = 2x$$. Then why can't we solve a separable differential equation just by taking derivative of both sides for instance in $$\dfrac{1}{\cos(y)} \cdot dy = \sin(x) \cdot dx$$

• $x^2~dx \neq 2x$ but $d(x^2) = 2x$. Commented Oct 25, 2020 at 18:10
• If $\cos y=0$ original DE becomes $y'=0$ Commented Oct 25, 2020 at 18:16
• @VIVID Thank you! And what does $x^2dx$ mean? Commented Oct 25, 2020 at 18:18
• @HasanÖzden In my previous comment, I omitted $dx$ in the end, so it must have been written: $d(x^2) = 2xdx$. About $dx$, better see: google.com/… Commented Oct 25, 2020 at 18:24

$$\frac{dy}{dx}=\sin x \cos y$$ Separate variables $$\frac{dy}{\cos y}=\sin x$$ Integrate both sides $$\int \frac{dy}{\cos y}=\int \sin x \,dx$$ Set $$\cos y=\frac{1-t^2}{1+t^2};\;t=\tan\frac{y}{2};\;y=2\arctan t;dy=\frac{2dt}{1+t^2}$$ $$\int \frac{dy}{\cos y}=\int \frac{2dt}{1+t^2}\cdot \frac{1+t^2}{1-t^2}=\int \frac{2dt}{1-t^2}=2 \text{arctanh}\, t+C=\\= 2 \text{arctanh}\, \tan\frac{y}{2}+C$$ So the solution can be written as $$2 \text{arctanh}\, \tan\frac{y}{2}=-\cos x + C\\ \text{arctanh}\, \tan\frac{y}{2}=-\frac{\cos x}{2} + \frac{C}{2}\\ \tan\frac{y}{2}=\tanh\left(\frac{\frac{C}{2}-\cos x}{2} \right)\\ y=2\arctan\left(\tanh \frac{1}{2}\left(c-\cos x \right)\right)$$

$$\int \frac{dy}{\cos y}=\int \sin x \,dx=-\cos x +C$$ Then for the integral on the LHS: $$I=\int \frac{dy}{\cos y}=\int \frac{d \sin y}{\cos^2 y}$$ $$I=\int \frac{d \sin y}{1-\sin^2 y}=\int \frac{d u}{1-u^2}$$ $$I=\dfrac 12 \left (\int \frac{du }{u+1}-\int \frac{du}{u-1}\right )$$ $$I=\dfrac 12 \ln \left |\dfrac {u+1}{u-1}\right |+C= \dfrac 12\ln \left |\dfrac {(u+1)^2}{u^2-1}\right |+C$$ Where $$u =\sin y$$. $$I= \ln \left |\dfrac {1+\sin y}{\cos y}\right |+C$$

You can also use this result: $$I=\dfrac 12 \ln \left |\dfrac {u+1}{u-1}\right |+C=\tanh^{-1} u+C$$ $$\implies I=\tanh^{-1} ( \sin y)+C$$ So that we have: $$\tanh^{-1} ( \sin y)=(C-\cos x)$$ $$y(x)=\arcsin (\tanh(C-\cos x))$$

$$\int\frac{dy}{\cos(y)}=\int\frac{\cos(y)\,dy}{1-\sin^2(y)}=\text{artanh}(\sin(y))=\int\sin (x)\,dx$$

gives

$$y=\arcsin(\tanh(c-\cos(x)))$$ and periodic replicas.

$$y=\pm\dfrac\pi2$$ and replicas are also (degenerate) solutions ($$y'=0$$, corresponding to $$c=\pm\infty$$).

• No, $\text{artanh}(y)$ is wrong. $\int \sec(y)\; dy = \ln(\sec(y)+\tan(y)) + C$. Commented Oct 25, 2020 at 18:17
• @RobertIsrael: yep, I dropped the sine. Now fixed. Thanks.
– user65203
Commented Oct 26, 2020 at 8:36