Solve $\dfrac{dy}{dx}=\cos(x+y)+\sin(x+y)$

Question

Solve $\frac{dy}{dx}=\cos(x+y)+\sin(x+y)$

My Attempt $$ \frac{dy}{dx}=\sqrt{2}\cos(x+y-\tfrac{\pi}{4}) $$ Set $t=x+y-\tfrac{\pi}{4}\implies y=t-x+\tfrac{\pi}{4}$ $$ \frac{dy}{dx}=\frac{dt}{dx}-1=\sqrt{2}\cos t\\ \frac{dt}{dx}=\sqrt{2}\cos(t)+1\\ \int\frac{dt}{\sqrt{2}\cos(t)+1}=\int dx $$ How do I proceed further to find the general solution ?

Answer 1

$$ \begin{align} &\int dx=\int\frac{dt}{\sqrt{2}\cos(t)+1}\\ x&=\frac{1}{\sqrt{2}}\int\frac{dt}{\cos t+\frac{1}{\sqrt{2}}}=\frac{1}{\sqrt{2}}\int\frac{dt}{\cos t+\cos\frac{\pi}{4}}=\frac{1}{\sqrt{2}}\int\frac{dt}{2\cos\big(\frac{t}{2}+\frac{\pi}{8}\big)\cos\big(\frac{t}{2}-\frac{\pi}{8}\big)}\\ &=\frac{\sqrt{2}}{2\sqrt{2}}\int\frac{\sin\Big[\big(\frac{t}{2}+\frac{\pi}{8}\big)-\big(\frac{t}{2}-\frac{\pi}{8}\big)\Big]}{\cos\big(\frac{t}{2}+\frac{\pi}{8}\big)\cos\big(\frac{t}{2}-\frac{\pi}{8}\big)}dt\\ &=\frac{1}{2}\bigg[\int\tan\Big(\frac{t}{2}+\frac{\pi}{8}\Big)dt-\int\tan\Big(\frac{t}{2}-\frac{\pi}{8}\Big)dt\bigg]\\ &\text{Set }u=\frac{t}{2}+\frac{\pi}{8}\implies dt=2du \quad\&\quad v=\frac{t}{2}-\frac{\pi}{8}\implies dt=2dv\\ x&=\int\tan u.du-\int\tan v.dv=\log|\sec u|-\log|\sec v|=\log|\frac{\sec u}{\sec v}|+C\\ &=\log|\frac{\cos v}{\cos u}|+C=\log\bigg|\frac{\cos\big(\tfrac{t}{2}-\tfrac{\pi}{8}\big)}{\cos\big(\tfrac{t}{2}+\tfrac{\pi}{8}\big)}\bigg|+C\\ &=\log\bigg|\frac{\cos\big(\tfrac{x+y}{2}-\tfrac{\pi}{8}-\tfrac{\pi}{8}\big)}{\cos\big(\tfrac{x+y}{2}-\tfrac{\pi}{8}+\tfrac{\pi}{8}\big)}\bigg|+C=\log\bigg|\frac{\cos\big(\tfrac{x+y}{2}-\tfrac{\pi}{4}\big)}{\cos\big(\tfrac{x+y}{2}\big)}\bigg|+C\\ &=\log\bigg|\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\tan\big(\tfrac{x+y}{2}\big)\bigg|+C\\ &\color{red}{x=\log\bigg|1+\tan\big(\tfrac{x+y}{2}\big)\bigg|+K} \end{align} $$

Answer 2

Making $x+y = u\to \frac{dy}{dx} = \frac{du}{dx}-1$

$$ \frac{du}{dx} = \sin u+ \cos u +1 $$

this is a separable DE

$$ \frac{du}{\sin u+\cos u + 1} = dx $$

etc.

NOTE

$$ \sin u+\cos u = \sqrt 2\sin(u+\frac{\pi}{4}) $$

now using the identities

$$ \cos u=2\cos^2(\frac u2)-1\\ \sin u =2\cos(\frac u2)\sin(\frac u2) $$

we have

$$ \frac{1}{\cos u+\sin u+1} = \frac{1}{2\cos(\frac u2)(\cos(\frac u2)+\sin(\frac u2))} = \frac 12\left(\frac{\sin(\frac u2)}{\cos(\frac u2)}+\frac{-\sin(\frac u2)+\cos(\frac u2)}{\cos(\frac u2)+\sin(\frac u2)}\right) $$

etc.