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Cauchy Problem, Same variables in one fraction

  1. Sep 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve the following Cauchy problem .

    y' = (3y2 + t2)/ 2ty
    y(1) = 1

    2. Relevant equations

    3. The attempt at a solution

    I know that the equation is not separable or linear from the get-go. I spoke with my professor and he said it is related to the squared variables, so I was thinking some sort of substitution, like z = f(y,t) and going from there, but I don't know how to actually do it. :(
  2. jcsd
  3. Sep 3, 2011 #2
    Try to check if it is an exact equation. (it isn't).
    But maybe there exists a suitable integration factor??
  4. Sep 3, 2011 #3
    (working on it now)
  5. Sep 3, 2011 #4
    It has to be in standard form first, so then it becomes

    (2ty)y' - 3y2 = t2

    y' - 3y/(2t) = t/(2y)

    Don't all the y's have to be on the left?
  6. Sep 3, 2011 #5
    What is the standard form of an exact ODE??


  7. Sep 3, 2011 #6
    y'(t,y) + p(t)y = g(t,y), I believe.
  8. Sep 4, 2011 #7


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    the point is that both numerator and denominator are of "degree 2". If you were to multiply both y and t by "a" on the right, the "a"s would all cancel out. This is a "homogenous" first order equation.

    [tex]y'= \frac{3y^2+ t^2}{2ty}= \frac{3}{2}\frac{y}{t}+ \frac{1}{2}\frac{t}{y}[/tex]

    Try the substitution u= y/t so that y= tu.
  9. Sep 4, 2011 #8
    I actually tried that before! Maybe I'm just doing it wrong, then.

    (working on it now)

    Here's what I got so far (kind of stuck)

    t (du/dt) = u/2 + 1/(2u)


    (1/t) dt = [4u / (2u2 + 2)] du

    ln|t| + C = ln|2u2+2|

    Not sure how to proceed from here. I'm assuming we can "say" C = 0? Or that, for some reason,

    ln(C)|t| = ln|t| + C.

    That's what my professor did for a similar example (still don't know why it works).

    And say that the insides of the ln are the same...?

    So then,

    C|t| = |2u2 + 2|

    C = |2u2 +2| / |t|

    What happens now?? :(
    Last edited: Sep 4, 2011
  10. Sep 4, 2011 #9
    First of all, [itex]4u / (2u^2 + 2) = 2u / (u^2 + 1)[/itex], which simplifies the anti-derivative slightly to [itex]ln|u^2+1|[/itex]. You can't make the assumption that [itex]C=0[/itex]; you have an initial condition that must be obeyed!

    Note that [itex]C = ln(e^C)[/itex]; then you can combine logarithms with the laws of logarithm addition and solve for [itex]u[/itex].
  11. Sep 5, 2011 #10
    Okay then,

    ln|t| + C = ln|2u2+2|

    I will now use

    ln|t| + C = ln|u2+1|

    Now, replacing the C with ln(eC),

    ln|t| + ln|eC| = ln|u2 + 1|

    (Working now)
  12. Sep 6, 2011 #11
    Just finished it up.

    I end up with:

    ln|t + eC| = ln|u2+1|

    Insides of ln must be equal,

    t + eC = u2 + 1

    u2 = t + eC - 1

    Since u = (y/t),

    (y/t)2 = t + eC - 1

    So: y = t (sqrt(t + eC - 1))

    Using y(1) = 1,

    C must equal zero.

    y = t(sqrt(t))

    Last edited: Sep 6, 2011
  13. Sep 6, 2011 #12
    Dude, [itex]ln(a) + ln(b) = ln(ab)[/itex]. You should have [itex]ln|t e^{c}| = ln|u^2 + 1|[/itex]. Also, don't forget the plus-or-minus sign when you take the square root of both sides. You'll need check which sign you want by which one works with the initial condition.
  14. Sep 6, 2011 #13
    Gosh. Dangit. I can't believe I was making that mistake. I really thought I multiplied the two... grr.

    Thanks for the point-out though.

    In this case, I got

    1 = 1 (+ or -) sqrt(eC - 1)

    In order for it to be positive, it should be plus rather than minus. But it doesn't really matter, because then sqrt(eC - 1) = 0 no matter what, right? Even if it were negative or positive, the C would have to equal zero for the sqrt of something to be zero.

    My final answer is now

    y = t (+ or -) sqrt(t - 1)

    Any thoughts?
  15. Sep 6, 2011 #14
    Okay, let's back up for a second. You presumably found that:

    [tex] y= \pm t \sqrt{te^c-1} [/tex]

    I'll show you a "cool" trick. Since [itex]e[/itex] is a constant, [itex]e^C[/itex] must also be a constant, so you can just replace it with [itex]C[/itex].

    [tex] y= \pm t \sqrt{Ct-1} [/tex]

    You sub in [itex]y(1) = 1[/itex]:

    [tex] 1 = \pm \sqrt{C-1}[/tex]

    Now do you see which sign we have to take?
  16. Sep 7, 2011 #15
    I'm thinking the sign should be positive, because if you fit in the initial condition into the negative form, then found C, then put it back into the negative form, the answer would have

    1 = - (1) * sqrt (2-1) which is an false statement...?
  17. Sep 7, 2011 #16

    Alternatively, you could note that since the left side is positive, the right side must also be positive, so we have to take the plus sign since the square root function is never negative.
  18. Sep 7, 2011 #17
    Awesome. Thanks so much!
  19. Sep 7, 2011 #18
    No prob :biggrin:
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