## cwrw238 2 years ago I'm having trouble solving this differential equation (1 - x^2) y' - xy = x I'm not sure about the integrating factor is it (1 - x^2) ^(1/2) or 1/ ((1 - x^2)^(1/2) (or maybe its neither!). Please help.

1. phi

if in the form y' + P(x) y = Q(x) the integrating factor is e^F where F= integral P(x) dx

2. SmartWish

$(1-x^2)\frac{dy}{dx} -xy = x$

3. cwrw238

@ph1 - yes - I remembered that I'm (to my shame) confuse by the algebra

4. phi

integral -x dx/(1-x^2) = 1/2 * ln(1-x^2) (toss in abs signs to be sure it's positive)

5. phi

exp (ln(1-x^2)^(1/2))= sqrt(1-x^2)

6. phi

integral du/u = ln(u) in this case u = 1-x^2 du = -2x dx

7. phi

$e^{ab} = \left(e^a\right)^b$ so $e^{\ln(1-x^2)\cdot \frac{1}{2}}=\left(e^{\ln(1-x^2)}\right)^\frac{1}{2}$ $= (1-x^2)^\frac{1}{2}= \sqrt{1-x^2}$

8. cwrw238

ok - i see it now . I was confused with the sign when integrating thanx

9. cwrw238

i need more practice in FODEs .......

10. phi

are you saying you can't finish ?

11. cwrw238

no i think i've got it now I get the solution to be y = A(1 - x^2)^(-1/2) - 1

12. phi

that looks good

13. cwrw238

I got it down to |dw:1359820446243:dw|

14. cwrw238

from here its easy thanks again

15. phi

I would expect to see $\sqrt{1-x^2}y = \int\limits \sqrt{1-x^2}\frac{x}{1-x^2} dx$ the integral gives $- \sqrt{1-x^2}+ A$ so $\sqrt{1-x^2}y = A - \sqrt{1-x^2}$ and finally $y= \frac{A}{\sqrt{1-x^2}} - 1$

16. phi

though you could have tossed in a minus sign on both sides before integrating ?

17. cwrw238

hmm i'm confused how come the integral is negative of square root bit?

18. cwrw238

i'm allergic to integration!!!

19. phi

$\int\limits \sqrt{1-x^2}\frac{x}{1-x^2} dx = \int (1-x^2)^{- \frac{1}{2}} x dx$ u is 1-x^2 and du = -2x dx we need a -2 so multiply and divide by -2: $- \frac{1}{2} \int (1-x^2)^{- \frac{1}{2}} -2x dx$ we now have $- \frac{1}{2} \int u^{- \frac{1}{2}} du= - \frac{1}{2} \cdot 2 u^{\frac{1}{2}}$

20. cwrw238

oh ok - I used the following standard form which probably isnt valid in this case Integral x dx / sqrt(a^2 + x^2) = sqrt (a^2 + x^2)

21. phi

except you have a^2 - x^2 not a^2 + x^2

22. phi

maybe the Greeks had the right idea, stick with positive rational numbers at all times

23. phi

then none of these stray minus signs to keep track of.

24. cwrw238

yep this is probably a silly question but since the left hand side is the result of the product rule shoudntt it be - y sqrt(1 - x^2) ?

25. cwrw238

yes its the signs which have confused me here

26. phi

The integrating factor is sqrt(1-x^2). It is positive. call the integrating factor F(x) and start with dy/dx + P(x) y = Q(x) F(x) * dy/dx + F(x)*P(x) y = F(x)*Q(x) and F(x) dy + P(x) y F(x) dx = F(x)Q(x) dx remember F(x) = exp(integral(P(x))) d F(x) = P(x) exp( integral(P(x))) dx we could work through the details, but it looks ok to me. we hae F(x) dy + y d (F(x)) = F(x)Q(x) dx or d( F(x) y) = F(x)Q(x) dx

27. phi

the "tricky part" is taking the derivative of $\frac{d}{dx} e^{\int P(x) dx} = e^{\int P(x) dx} \cdot \frac{d}{dx}\int P(x) dx= e^{\int P(x) dx} \cdot P(x) dx$

28. cwrw238

yes ok i'll have a good look at this - I'm still a bit mistifi ed (maybe its because i've got a cold!) ty

29. phi

if this was easy, anybody could do it.

30. cwrw238

right