Integrating factor!!


by james.farrow
Tags: factor, integrating
james.farrow
james.farrow is offline
#1
Feb16-10, 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??
Phys.Org News Partner Science news on Phys.org
Lemurs match scent of a friend to sound of her voice
Repeated self-healing now possible in composite materials
'Heartbleed' fix may slow Web performance
james.farrow
james.farrow is offline
#2
Feb16-10, 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!!!
vela
vela is offline
#3
Feb16-10, 12:21 PM
Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,521
That's almost right. Did you try plugging it back into the original equation to see if it worked?

james.farrow
james.farrow is offline
#4
Feb16-10, 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
Mark44
Mark44 is online now
#5
Feb16-10, 02:40 PM
Mentor
P: 20,933
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.
james.farrow
james.farrow is offline
#6
Feb16-10, 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
Mark44
Mark44 is online now
#7
Feb16-10, 03:08 PM
Mentor
P: 20,933
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.
james.farrow
james.farrow is offline
#8
Feb16-10, 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?
Mark44
Mark44 is online now
#9
Feb16-10, 03:33 PM
Mentor
P: 20,933
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.
james.farrow
james.farrow is offline
#10
Feb16-10, 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
vela
vela is offline
#11
Feb16-10, 03:38 PM
Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,521
Hint: Your mistake has to do with when you introduced the constant of integration.
james.farrow
james.farrow is offline
#12
Feb16-10, 03:40 PM
P: 44
Hmmm I'm not sure, but at a guess should it be ln(C) not just C ??
james.farrow
james.farrow is offline
#13
Feb16-10, 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
vela
vela is offline
#14
Feb16-10, 03:51 PM
Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,521
Yup, that's it. You can, of course, check your answer by plugging it back into the original differential equation.
james.farrow
james.farrow is offline
#15
Feb16-10, 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