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Solving the DE

  1. May 13, 2008 #1
    Hello!

    Here's a question I was doing but I'm not sure if I'm right...

    It says "solve" the following differential equation:
    [tex](x^2 +1) \frac{dy}{dx} = 0[/tex]

    So, this is what I've done:

    We put it in form [tex]\frac{dy}{dx} +p(x)y = q(x)[/tex]

    Then we devide it by x^2 +1 to obtain:
    [tex]\frac{dy}{dx} + \frac{x}{x^2 +1} y = \frac{1}{x^2 +1} \times 0[/tex]

    you know , in standard form where p(x) = x/x^2 +1

    Integrating factor is:
    [tex]exp[\int p(x) dx = exp[\frac{x dx}{x^2 +1}[/tex]
    [tex]= exp[ln(x^2 +1)^{2}][/tex]
    [tex]= (x^2 +1)^{2}[/tex]

    Multiplying the DE by the integrating factor;
    [tex](x^2 +1)^{2} \frac{dy}{dx} +x(x^2 +1)y = x . (x^2 +1)^{2}[/tex]

    Now we integrate:
    [tex]\frac{d}{dx} [(x^2 +1)^{2} y] = x^3 +x[/tex]

    => [tex][(x^2 +1)^{2} y] = \frac{x^4}{4} + \frac{x^2}{2} + c [/tex]

    Am I right? Please correct me if I'm wrong...

    Thank you,
     
  2. jcsd
  3. May 13, 2008 #2

    Vid

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    The only answer is y = c.

    You have two things being multiplied together equaling zero. This means that either (x^2+1) = 0 or dy/dx = 0. In the first case, the equation is only true when x = +-i which isn't very useful since we solve DEs over an interval and would also allow for y to be any function of x (again not very useful). In the second case, dy/dx = 0 thus y = c is the only solution.

    You can see your solution doesn't work by just plugging it into the original equation.
     
  4. May 13, 2008 #3
    Yes, maybe but could you please show me how the actual calculation goes till we get to y = c? :rolleyes:

    Thank you.
     
    Last edited: May 13, 2008
  5. May 13, 2008 #4

    Vid

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    dy/dx = 0 from my explanation above. The only (continuous) functions with zero derivatives are constant functions.
     
  6. May 14, 2008 #5
    [tex](x^2 +1) \frac{dy}{dx} = 0[/tex]


    The correct answer at the back of the book is [tex]y = \frac{c}{\sqrt{x^2 +1}}[/tex].

    I get your explaination, Vid. But the question is asking me to show working and calculations.
     
    Last edited: May 14, 2008
  7. May 14, 2008 #6

    HallsofIvy

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    No, you haven't understood what he is saying. Vid is trying to tell you is that, since x2+ 1 is never 0, you can divide on both sides by it to get dy/dx= 0 which, integrating, gives y= c, a constant. That should be pretty simple mathematics.
    [itex]y(x)= c\sqrt{x^2+1}[/itex] satisfies the equation [itex](x^2+1)dy/dx= xy[/itex], not 0.
     
  8. May 14, 2008 #7
    Yes, Thanks. :smile:

    2) Solve: [tex]\frac{dy}{dx} + 2xy = x[/tex]

    This is what I think;

    [tex]\int 2x dx = x^2[/tex]

    So we multiply both sides by [tex]e^{x^2}[/tex]

    [tex]\frac{d}{dx} (e^{x^2} y) = xe^{x^2}[/tex]

    [tex]e^{x^2}y = \frac{1}{2}e^{x^2} + c[/tex]

    => [tex]y = c e^{x^2} - \frac{1}{2}[/tex]

    c is a constant.

    I appreciate that if you could tell me if I made any errors or mistakes..
     
    Last edited: May 14, 2008
  9. May 15, 2008 #8

    Defennder

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    It's correct up to here.

    This is incorrect. You were dividing the RHS by [tex]e^{x^2}[/tex] right? You should get [tex]\frac{1}{2} + ce^{-x^2}[/tex] instead.
     
  10. May 15, 2008 #9
    [tex]\frac{1}{2} + ce^{-x^2}[/tex] or [tex]\frac{1}{2} + ce^{x^2}[/tex]?

    Anyway, Thanks a lot for your attention!
     
  11. May 15, 2008 #10

    Defennder

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    It's the first one. That is what you get when you divide c by [tex]e^{x^2}[/tex]
     
  12. May 15, 2008 #11
    Of course!
    Many thanks...
     
  13. May 15, 2008 #12
    For the first question, if the question was: [tex](x^2 +1) \frac{dy}{dx} +xy = 0[/tex]
    then the solution would be: [tex]y = \frac{c}{\sqrt{x^2 +1}}[/tex]

    Hmm, so how do we get that? (that was my question but I missed out +xy).

    Does any of my working in the very above post make any sense?

    I don't know how do this question so I get [tex]y = \frac{c}{\sqrt{x^2 +1}}[/tex] ...
     
    Last edited: May 15, 2008
  14. May 15, 2008 #13

    Defennder

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    What's the integrating factor in this case? Remember you must first express the DE in the form: [tex]\frac{dy}{dx} + P(x)y = Q(x)[/tex]

    Your answer appears to be correct, so what's the problem?
     
  15. May 15, 2008 #14

    HallsofIvy

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    So [tex]\frac{dy}{dx}= x- 2xy= x(1- 2y)[/itex]
    [tex]\frac{dy}{1- 2y}= xdx[/itex]

     
  16. May 15, 2008 #15
    Yes I did put it in the standard form and I found the integrating factor

    [tex](x^2 +1) \frac{dy}{dx} +xy = 0[/tex]

    standard form : [tex]\frac{dy}{dx} +p(x)y = q(x)[/tex]

    Then we devide it by x^2 +1 to obtain:
    [tex]\frac{dy}{dx} + \frac{x}{x^2 +1} y = \frac{1}{x^2 +1} \times 0[/tex]

    Integrating factor is:
    [tex]exp[\int p(x) dx = exp[\frac{x dx}{x^2 +1}[/tex]

    [tex]= exp[ln(x^2 +1)^{2}][/tex]
    [tex]= (x^2 +1)^{2}[/tex]

    multiplying it by the integrating factor;
    [tex](x^2 +1)^{2} \frac{dy}{dx} +x(x^2 +1)y = x . (x^2 +1)^{2}[/tex]

    Now we integrate: [tex]\frac{d}{dx} [(x^2 +1)^{2} y] = x^3 +x[/tex]


    I get: [tex][(x^2 +1)^{2} y] = \frac{x^4}{4} + \frac{x^2}{2} + c [/tex]

    But this is wrong since the correct answer should be: [tex]y = \frac{c}{\sqrt{x^2 +1}}[/tex]

    I don't know how to get that.:cry:
     
  17. May 15, 2008 #16

    Vid

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    Int(x/(x^2+1) = 1/2ln(x^2+1) = ln|(x^2+1)^(1/2)| not squared. The right side is just zero so the integral is just c; solve for y to get the right answer. Also this DE is easier as a separable equation instead of a first order linear.
     
    Last edited: May 15, 2008
  18. May 16, 2008 #17

    HallsofIvy

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    There is no need to use integrating factors: you have a separable equation.
    [tex](x^2 +1) \frac{dy}{dx} +xy = 0[/tex]
    Subtractiong xy from both sides and dividing on both sides by [itex]x^2+ 1[/itex]
    [tex]frac{dy}{dx}= \frac{xy}{x^2+ 1}= \right(\frac{x}{x^2+1}\right) y[/tex]
    [tex]ydy= \frac{x dx}{x^2+1}[/tex]
     
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