Solve the differential equation,$$\frac{\mathrm{d}y}{\mathrm{d}x} =\cos(x+y) + \sin(x+y)$$

My attempt at the problem:

$$\frac{\mathrm{d}y}{\mathrm{d}x} =\cos(x+y) + \sin(x+y)=\cos(x+y)\biggr(1+\tan(x+y)\biggr)$$

Now let, $$\cos(x+y)=u$$

So that,

$$\frac{\mathrm{d}y}{\mathrm{d}x} = u(1+\sqrt{\frac{1}{u^2}-1})$$ using $\tan^2A = \sec^2A - 1$


$$\frac{\mathrm{d}(\cos(x+y))}{\mathrm{d}x} = -\sin(x+y)\biggr(1 + \frac{\mathrm{d}y}{\mathrm{d}x} \biggr) $$

and $$\frac{\mathrm{d}u}{\mathrm{d}x} = -\sqrt{1-u^2}\biggr(1 + \frac{\mathrm{d}y}{\mathrm{d}x} \biggr)$$

Solving for $\frac{\mathrm{d}y}{\mathrm{d}x}$ and plugging it in the problem gives a very ugly integral upon separating variables

$$\int{\mathrm{d}x} = - \int \frac{\mathrm{d}u}{\biggr((u+1)+\sqrt{1-u^2}\biggr)\sqrt{1-u^2}}$$

I just can't solve it any further.

Please correct me if I'm wrong or please direct me towards an alternate solution. All help appreciated!


1 Answer 1


set $$x+y=u$$ then we have $$y=u-x$$ and $$y'=u'-1$$ and we have $$u'-1=\cos(u)+\sin(u)$$ Can you solve this?

  • $\begingroup$ you have made $$\cos(x+y)=u$$ and i have Substitute $$x+y=u$$ there is a difference $\endgroup$ Dec 6, 2017 at 16:57
  • $\begingroup$ Can you solve it, Dr. Graubner? $\endgroup$ Dec 6, 2017 at 17:27
  • 1
    $\begingroup$ @TobErnack The left-hand side is $u'-1$ and not $u'$, right? $\endgroup$
    – mickep
    Dec 6, 2017 at 18:46
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    $\begingroup$ @TobErnack Indeed, but the primitive actually looks simpler, since $\int 1/(1+\cos u+\sin u)\,du$ becomes $\ln(1+\tan(u/2))$ or something similar. $\endgroup$
    – mickep
    Dec 6, 2017 at 19:46
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    $\begingroup$ Since the only missing piece from the excellent answer of @Dr.SonnhardGraubner is the $t=\tan(u/2)$ substitution, I suggest that I do not add another answer with only that detail. $\endgroup$
    – mickep
    Dec 7, 2017 at 6:28

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