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Differential equations

  1. Feb 14, 2010 #1
    I have the following to problem:-

    1)dy/dt = (1+y^2)/(1+4t^2) y(0)=1

    This equation is seperable I arrive at

    Integral 1/(1+y^2) dy = Integral 1/(1+4t^2) dt

    If I let 2t=x then the RHS is in the same form as my standard integrals.

    Which is Integral 1/(x^2 + a^2) = 1/a arctan(x/a)

    So far so good methinks...

    After a bit of 'jiggery pokery' I get

    arctan(y) = arctan(2t) + C

    Take tan of both sides -> y = 2t + tan(C)
    Putting in initial values

    1 = 2*0 + tan(C)

    Therefore C = pi/4

    Finally y = 2t + pi/4 -> Am I right!!

    Now my second equation is thus:

    (x^2 + 1)dy/dx -2xy = 2x(x^2 +1)

    Putting it into standard from

    dy/dx - (2xy)/(x^2 + 1) = 2x

    This gives me an integrating factor 1/(x^2 +1)

    Now what!?
  2. jcsd
  3. Feb 14, 2010 #2
    Methinks this step is an error. Note the following identity.

    [tex]{\rm \tan}(a+b)={{{\rm tan}a+{\rm tan}b}\over{1-{\rm tan}a\; {\rm tan}b}}[/tex]
  4. Feb 14, 2010 #3


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    You can check your answer by plugging y into the formula.
    However, you might want to check this step:
    because tan(a + b) [itex]\neq[/itex] tan(a) + tan(b) in general.
  5. Feb 14, 2010 #4
    Cheers everyone! If I use the identity I still get C = pi/4.

    What is the check??

  6. Feb 14, 2010 #5


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    You dropped a factor of 1/2 when integrating the RHS.
  7. Feb 15, 2010 #6
    Can you show me how I dropped the factor of 1/2?

    I have 1/(4t^2 + 1)

    If x=2t then x^2=4t^2

    So using 1/(x^2 + a^2) where a=1=a^2

    I'm struggling with it really!!

    Cheers James
  8. Feb 15, 2010 #7
    Ok everyone I've had a rethink on my integration...

    Here goes.

    Integral of 1/(1 + 4t^2) becomes

    1/4 Integral of 1/(1/4 + t^2)

    a^2= 1/4 so a=1/2

    Standard integral 1/a arctan(x/a)

    So finally after integrating I arrive at

    1/2 arctan(2t)

    What do you think?

  9. Feb 15, 2010 #8


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    That's correct.

    In your previous attempt, you have to replace dt by dx. Because x=2t, you had dx=2 dt, or dt=dx/2. That's where your missing factor of 1/2 went.
  10. Feb 15, 2010 #9
    Thanks for your help everyone! It's appreciated!

    No doubt you'll be hearing from me shortly - integrating factor is next...
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