Proof of Integration Factor for diff eq

Click For Summary
SUMMARY

This discussion focuses on proving the existence of an integrating factor for the first-order differential equation represented by M(t,y) + N(t,y)dy/dt = 0. The integrating factor is defined as μ(y) = exp(integral Q(y)dy), where Q(y) is derived from the condition ((partial M/partial t) - (partial N/partial y))/M = Q(y). Participants emphasized the importance of verifying exactness through partial derivatives and the correct application of the chain rule during differentiation.

PREREQUISITES
  • Understanding of first-order differential equations
  • Familiarity with integrating factors in differential equations
  • Knowledge of partial derivatives
  • Proficiency in applying the chain rule in calculus
NEXT STEPS
  • Study the derivation of integrating factors for different types of differential equations
  • Learn about the exactness condition in differential equations
  • Explore the application of the chain rule in multivariable calculus
  • Investigate the implications of changing limits of integration in definite integrals
USEFUL FOR

Students and educators in mathematics, particularly those focused on differential equations, as well as researchers and practitioners needing to apply integrating factors in their work.

Ald
Messages
24
Reaction score
1

Homework Statement



Looking for help on this Exact first order differential equation problem. Thanks

Show that if ((partial M/partial t)-(partial N/partial y))/M=Q(y)

then the differential equation M(t,y)+N(t,y)dy/dt=0 has an integrating factor

μ(y)=exp(integral Q(y)dy).



Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
If mu(y) is an integrating factor, then mu(y)*M(t,y)*dt+mu(y)*N(t,y)*dy=0 is exact. Is it? Try the usual test for exactness. BTW, be clear where the y dependence is in mu(y). It's not in the integration variable, that's just a dummy. y is one of the limits of integration.
 
Thanks for your help, yes your recommendation makes sense.

I assume these are the partial derivatives I take, there is no actual expression to take the partial of, how do I prove exactness?

partial(mu(y)*M(t,y))/partial dy=partial (mu(y)*N(t,y))/partial dt
 
Use your formula for mu(y). The partial derivative of mu(y) wrt to y can be written in a different way. The partial derivatives of M and N, you just write as partial derivatives. You are trying to show that mu is an integrating factor leads to the expression in the 'if' part of your statement.
 
I'm sorry what does 'wrt' mean?
 
wrt='with respect to'. What's the derivative of mu(y) with respect to y?
 
The only other way I could find to write the mu(y) was in the form mu(t,y)=1/(xM-yN) is this what you had in mind?
Thanks
 
Thanks Dick for getting me in right direction, I think I have it proved.
Thanks again
 
Ald said:
Thanks Dick for getting me in right direction, I think I have it proved.
Thanks again

Did it really work? You differentiated the exponential and used the chain rule, right? I'm kind of curious. I had to change e^{\int_0^y Q(t)dt} by putting a negative sign inside the exp. You could also alter it by setting y as the lower limit, of course. Did you hit that snag?
 
  • #10
Attached is what I did, I haven't rewritten yet. Sometimes ignorance is bliss, I hope I did it right.
 

Attachments

  • #11
If ignorance is bliss, I'm experiencing it now. An attachment can take hours to clear approval.
 

Similar threads

Replies
3
Views
2K
Replies
12
Views
2K
Replies
7
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K
Replies
6
Views
2K
Replies
24
Views
3K