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 $\dfrac{1}{\cos(y)} \cdot dy = \sin(x) \cdot dx$

  • 1
    $\begingroup$ $x^2~dx \neq 2x$ but $d(x^2) = 2x$. $\endgroup$
    – VIVID
    Commented Oct 25, 2020 at 18:10
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    $\begingroup$ If $\cos y=0$ original DE becomes $y'=0$ $\endgroup$
    – Raffaele
    Commented Oct 25, 2020 at 18:16
  • $\begingroup$ @VIVID Thank you! And what does $x^2dx$ mean? $\endgroup$
    – Hasan
    Commented Oct 25, 2020 at 18:18
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    $\begingroup$ @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/… $\endgroup$
    – VIVID
    Commented Oct 25, 2020 at 18:24

3 Answers 3


$$\frac{dy}{dx}=\sin x \cos y$$ 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: $$\tanh^{-1} ( \sin y)=(C-\cos x)$$ $$ y(x)=\arcsin (\tanh(C-\cos x))$$


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


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

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

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

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