Answer to Question #288268 in Differential Equations for Kiah

Question

y'=cos(x+y)

Answer 1

Instruction:

Solve $y'=cos(x+y)$

Solution:

$y'=cos(x+y) ...(1)$

Let $x+y = t$ ...(2)

$1+y'=\frac{dt}{dx}\Rightarrow y'=\frac{dt}{dx}-1$

From (1), we get

$\frac{dt}{dx}-1=\cos t\\\Rightarrow \frac{dt}{dx}=1+\cos t\\\Rightarrow \frac{dt}{dx}=2\cos^2 \frac{t}{2}\\ \Rightarrow \sec^2\frac{t}{2} dt=2dx$

Integrating both sides, we get

$2\tan \frac{t}{2}=2x+c\\ \Rightarrow 2\tan \frac{x+y}{2}=2x+c\\$ [From (2)]