How to Solve cos(x+y)dy=dx Using Trigonometric Identities?

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Discussion Overview

The discussion revolves around solving the differential equation cos(x+y)dy=dx using trigonometric identities and substitutions. Participants explore various methods to manipulate the equation, including the use of substitutions and transformations to separate variables.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests letting t = x+y to simplify the equation but expresses difficulty in separating variables.
  • Another participant proposes considering dx/dy, transforming the equation into (1 + cos(t)) = dt/dy.
  • A different viewpoint argues that the equation can indeed be made separable, detailing the steps to rewrite cos(x+y)dy=dx in terms of t and separating the variables accordingly.
  • Another participant presents a method involving u = x+y and derives a relationship for dy/dx, leading to a different expression for the integral.

Areas of Agreement / Disagreement

There is no consensus on whether the equation is separable. Some participants assert it can be separated while others express uncertainty or provide alternative methods without agreeing on a single approach.

Contextual Notes

Participants' methods depend on specific substitutions and transformations, which may introduce limitations based on the definitions and assumptions made during the discussion.

abrowaqas
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cos(x+y)dy=dx

I have tried to solve it by
Letting
t = x+y

But it's not going to be separated.. Some one please help..
 
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Consider dx/dy. The equation becomes (1 +cost)=dt/dy.
 
I don't see why you would say it does not become separable. If t= x+ y then dt= dx+ dy so dy= dt- dx. cos(x+y)dy= dx becomes cos(t)(dt- dx)= cos(t)dt- cos(t)dx= dx or cos(t)dt= (1+ cos(t))dx.
[tex]dx= \frac{cos(t)}{1+ cos(t)}dt[/tex].
 
I got the answer ..
Let u=x+y
du/dx=1+dy/dx
dy/dx=du/dx-1
Applying this:
cos(x+y)dy=dx
cos(x+y)dy/dx=1
cosu(du/dx-1)=1
du/dx-1=sec u
du/dx=1+sec u
dx/du=1/(1+sec u)
x=integral of 1/(1+sec u) du
 

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