Elementary Differential equations : seperable

  • Thread starter Mdhiggenz
  • Start date
  • #1
327
1

Homework Statement



Tan∅ dr + 2r d∅=0

Homework Equations





The Attempt at a Solution



∫1/2r dr +∫1/tan∅ d∅=0

1/2ln(2r)+ln(tan∅)=c
ln[r(tan∅)]=ln(c)

r(tan∅)=c

the solution in the book says the answer is rsin^2∅=c

Where did I go wrong?

Thank you
 

Answers and Replies

  • #2
3,816
92
Here's your mistake:
∫1/tan∅ d∅≠ln(tan∅)
 
  • #3
327
1
Here's your mistake:
∫1/tan∅ d∅≠ln(tan∅)
Wow I see it would be ∫cos∅/sin∅ d∅ and with some u substitution it would give us ln(sin∅) correct?
 
  • #4
3,816
92
Wow I see it would be ∫cos∅/sin∅ d∅ and with some u substitution it would give us ln(sin∅) correct?

Yep, that's right but you have done one more mistake.
∫1/2r dr≠(1/2)ln(2r). It is equal to (1/2)ln(r).
 

Related Threads on Elementary Differential equations : seperable

Replies
2
Views
1K
Replies
2
Views
2K
  • Last Post
Replies
0
Views
885
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
9
Views
3K
Replies
2
Views
1K
Replies
4
Views
1K
Top