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Integrating factor 
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#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

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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

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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

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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

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P: 21,214

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

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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

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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... 


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