# Integrating factor

1. Feb 16, 2010

### james.farrow

Integrating factor!!

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. Feb 16, 2010

### james.farrow

Re: Integrating factor!!

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. Feb 16, 2010

### vela

Staff Emeritus
Re: Integrating factor!!

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

4. Feb 16, 2010

### james.farrow

Re: 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. Feb 16, 2010

### Staff: Mentor

Re: Integrating factor!!

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. Feb 16, 2010

### james.farrow

Re: Integrating factor!!

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. Feb 16, 2010

### Staff: Mentor

Re: Integrating factor!!

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. Feb 16, 2010

### james.farrow

Re: Integrating factor!!

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. Feb 16, 2010

### Staff: Mentor

Re: Integrating factor!!

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. Feb 16, 2010

### james.farrow

Re: Integrating factor!!

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. Feb 16, 2010

### vela

Staff Emeritus
Re: Integrating factor!!

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

12. Feb 16, 2010

### james.farrow

Re: Integrating factor!!

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

13. Feb 16, 2010

### james.farrow

Re: Integrating factor!!

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. Feb 16, 2010

### vela

Staff Emeritus
Re: Integrating factor!!

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

15. Feb 16, 2010

### james.farrow

Re: Integrating factor!!

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...