Initial value problem - differential equations

  • #1
Cocoleia
295
4

Homework Statement


I am given (y^2 + y sin x cos y) dx + (xy + y cos x sin y) dy = 0, y(0) = π/2 .
I need to solve this

Homework Equations




The Attempt at a Solution


upload_2016-12-6_20-16-41.png


At this point they still aren't exact, so I gave up. I can't figure out what the problem is. Is it possible that I have to continue trying to make them exact?

I know that the answer is: xy-cosxcosy=0
 

Answers and Replies

  • #2
36,422
8,402

Homework Statement


I am given (y^2 + y sin x cos y) dx + (xy + y cos x sin y) dy = 0, y(0) = π/2 .
I need to solve this

Homework Equations




The Attempt at a Solution


View attachment 110019

At this point they still aren't exact, so I gave up. I can't figure out what the problem is. Is it possible that I have to continue trying to make them exact?

I know that the answer is: xy-cosxcosy=0
Divide both sides by y, to get ##(y + \sin(x)\cos(y))dx + (x + \cos(x)\sin(y)) dy = 0##.
Now check for exactness. Note that since ##y(0) = \pi/2##, it's reasonable to assume that ##y \ne 0##.
 
  • #3
Cocoleia
295
4
Divide both sides by y, to get ##(y + \sin(x)\cos(y))dx + (x + \cos(x)\sin(y)) dy = 0##.
Now check for exactness. Note that since ##y(0) = \pi/2##, it's reasonable to assume that ##y \ne 0##.
Can we always do this if we have a common factor like y in this case?
Thank you, I spent a very long time trying to figure this out
 
  • #4
36,422
8,402
Can we always do this if we have a common factor like y in this case?
I wouldn't say you can always do it. By getting rid of a factor of y, we are possibly dividing by zero. Since the initial condition is that ##y(0) = \pi/2##, if necessary, we can restrict y to some interval that includes ##\pi/2## but doesn't include zero.
 

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