## cwrw238 Group Title 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. one year ago one year ago

1. phi Group Title

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

2. SmartWish Group Title

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

3. cwrw238 Group Title

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

4. phi Group Title

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

5. phi Group Title

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

6. phi Group Title

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

7. phi Group Title

$e^{ab} = \left(e^a\right)^b$ 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 Group Title

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

9. cwrw238 Group Title

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

10. phi Group Title

are you saying you can't finish ?

11. cwrw238 Group Title

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

12. phi Group Title

that looks good

13. cwrw238 Group Title

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

14. cwrw238 Group Title

from here its easy thanks again

15. phi Group Title

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 Group Title

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

17. cwrw238 Group Title

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

18. cwrw238 Group Title

i'm allergic to integration!!!

19. phi Group Title

$\int\limits \sqrt{1-x^2}\frac{x}{1-x^2} dx = \int (1-x^2)^{- \frac{1}{2}} x dx$ 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 Group Title

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 Group Title

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

22. phi Group Title

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

23. phi Group Title

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

24. cwrw238 Group Title

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 Group Title

yes its the signs which have confused me here

26. phi Group Title

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 Group Title

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 Group Title

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 Group Title

if this was easy, anybody could do it.

30. cwrw238 Group Title

right