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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.

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  1. phi
    • 2 years ago
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    if in the form y' + P(x) y = Q(x) the integrating factor is e^F where F= integral P(x) dx

  2. SmartWish
    • 2 years ago
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    \[(1-x^2)\frac{dy}{dx} -xy = x\]

  3. cwrw238
    • 2 years ago
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    @ph1 - yes - I remembered that I'm (to my shame) confuse by the algebra

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

  5. phi
    • 2 years ago
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    exp (ln(1-x^2)^(1/2))= sqrt(1-x^2)

  6. phi
    • 2 years ago
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    integral du/u = ln(u) in this case u = 1-x^2 du = -2x dx

  7. phi
    • 2 years ago
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    \[ e^{ab} = \left(e^a\right)^b \] 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
    • 2 years ago
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    ok - i see it now . I was confused with the sign when integrating thanx

  9. cwrw238
    • 2 years ago
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    i need more practice in FODEs .......

  10. phi
    • 2 years ago
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    are you saying you can't finish ?

  11. cwrw238
    • 2 years ago
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    no i think i've got it now I get the solution to be y = A(1 - x^2)^(-1/2) - 1

  12. phi
    • 2 years ago
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    that looks good

  13. cwrw238
    • 2 years ago
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    I got it down to |dw:1359820446243:dw|

  14. cwrw238
    • 2 years ago
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    from here its easy thanks again

  15. phi
    • 2 years ago
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    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
    • 2 years ago
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    though you could have tossed in a minus sign on both sides before integrating ?

  17. cwrw238
    • 2 years ago
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    hmm i'm confused how come the integral is negative of square root bit?

  18. cwrw238
    • 2 years ago
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    i'm allergic to integration!!!

  19. phi
    • 2 years ago
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    \[\int\limits \sqrt{1-x^2}\frac{x}{1-x^2} dx = \int (1-x^2)^{- \frac{1}{2}} x dx\] 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
    • 2 years ago
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    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
    • 2 years ago
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    except you have a^2 - x^2 not a^2 + x^2

  22. phi
    • 2 years ago
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    maybe the Greeks had the right idea, stick with positive rational numbers at all times

  23. phi
    • 2 years ago
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    then none of these stray minus signs to keep track of.

  24. cwrw238
    • 2 years ago
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    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
    • 2 years ago
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    yes its the signs which have confused me here

  26. phi
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    if this was easy, anybody could do it.

  30. cwrw238
    • 2 years ago
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    right

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