Register to reply 
Integrating factor 
Share this thread: 
#1
Feb1610, 08:07 AM

P: 44

As promised I'm back with integrating factor differential equation.
(x^2 + 1)dy/dx 2xy = 2x(x^2+1) y(0)=1 First put into standard from by dividing thru by (x^2 +1 ) dy/dx 2xy/(x^2 + 1) = 2x Integrating factor is given by exp( integral of 2x(x^2 + 1)) After some working out I get the IF to be 1/(x^2 + 1) Now the solution is given by y(x)=1/IF(integral of 2x(x^2 + 1) Hopefully I'm on the right track so far... After doing the integration by parts and some tidying up I have y(x)= (x^2 + 1){(2(x^2 + 1)x^3)/3  4x^5/15} + C After plugging in the values I have 1=C What do you think?? 


#2
Feb1610, 08:20 AM

P: 44

On checking my work I think I've mad a mistake
the equation should be y(x) = 1/IF{integral of 2x/(x^2 + 1)} Which makes it different... My final revised answer is y(x) = (x^2 + 1)(ln{x^2 + 1}) + C After plugging in values y(0)=1 I have y(x)=(x^2 + 1)(ln{x^2 + 1}) + 1 Cheers!!! 


#3
Feb1610, 12:21 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,744

That's almost right. Did you try plugging it back into the original equation to see if it worked?



#4
Feb1610, 02:28 PM

P: 44

Integrating factor
I haven't and I'm not really sure how to do it or what you mean! Forgive my ignorance but can you show me....
James 


#5
Feb1610, 02:40 PM

Mentor
P: 21,280

You have your differential equation back in your original post in this thread. In post 2 you have a solution, y(x)=(x^2 + 1)(ln{x^2 + 1}) + 1.
Does this function satisfy the initial condition? I.e., is y(0) = 1? Does this function satisfy the differential equation? I.e., if you replace y and dy/dx in the differential equation with the function above and its derivative, do you get a true statement in this equation: (x^2 + 1)dy/dx 2xy = 2x(x^2+1)? You should always check your solutions to differential equations. 


#6
Feb1610, 02:46 PM

P: 44

Yes my solution satisfies condition y(0)=1
So I now need to differntiate my solution and substitute back into the equation? I'm not sure I get what you mean sorry? Bear in mind I'm trying to learn this...! lol James 


#7
Feb1610, 03:08 PM

Mentor
P: 21,280

Yes, that's what I mean. Take the derivative of your solution. Multiply it (the derivative) by (x^2 + 1). Subtract 2x times your solution. If you get 2x(x^2 + 1), your solution satisifies the DE.



#8
Feb1610, 03:19 PM

P: 44

I don't! After doing what you said I arrive at (x^2 + 1)  1
Which is x^2. So my solution is wrong? 


#9
Feb1610, 03:33 PM

Mentor
P: 21,280

Yes, it's wrong. That's what vela was suggesting that you do back in post #3. Now that you know you have a mistake, go back and take another look at your work and see if you can spot an error.



#10
Feb1610, 03:36 PM

P: 44

Thanks for your help Mark, I've been over my solution several times but always get the same  and its worng? I just can't see where Ive gone wrong...
Is it my integrating factor? James 


#11
Feb1610, 03:38 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,744

Hint: Your mistake has to do with when you introduced the constant of integration.



#12
Feb1610, 03:40 PM

P: 44

Hmmm I'm not sure, but at a guess should it be ln(C) not just C ??



#13
Feb1610, 03:45 PM

P: 44

Hold on! I think I may have it...?
I should have multiplied at all by 1/IF making my constant thus C(x^2 + 1) Or am I way off again... James 


#14
Feb1610, 03:51 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,744

Yup, that's it. You can, of course, check your answer by plugging it back into the original differential equation.



#15
Feb1610, 03:55 PM

P: 44

Thanks for your help everyone, my revised solution works! I've learned a lot, your gentle pointers eventually made the penny drop.
At least now I know how to check my solutions! Thanks again. James P.S I'll be moving onto 2nd order differential equations next and looking forward to your help again... 


Register to reply 
Related Discussions  
Help with integrating factor  Calculus & Beyond Homework  5  
Integrating factor  Calculus & Beyond Homework  3  
Integrating Factor  Calculus & Beyond Homework  2  
Integrating factor  Calculus & Beyond Homework  3  
Integrating factor  Introductory Physics Homework  2 