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Homework Help: Differential equations

  1. Jun 20, 2011 #1

    1. The problem statement, all variables and given/known data

    [tex] y'=(y-x)^{2}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I have nearly no idea in order to solve this solution. One hint is given: We shall find a new variable which makes it possible to seperate the differential equation afterwards.
    I'm not sure what's meant by "new variable". Shall I just find something cmpletly new, like let's call it z, or an addition like +x or *x etc.(just for example, I know this specific example doesn't work)?

    Thank you for your help
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 20, 2011 #2
    First thing I can suggest is posting this in the right place. Differential equations are definitely NOT precalculus
  4. Jun 20, 2011 #3


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    Gold Member

    Have you studied homogeneous functions? Your right side is homogeneous of degree 2 so try changing the dependent variable from y to u by y = ux.

    [Edit:] Never mind this suggestion. When the equation is put in the form Mdx + Ndy = 0 not both M and N are homogeneous.
    Last edited: Jun 20, 2011
  5. Jun 20, 2011 #4


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    Since it is the fact that it is y- x that is squared and not just x or y that is the problem, let z= y- x. Then z'= y'- 1 so that y'= z'+ 1= z^2. z'= z^2- 1.
    Last edited by a moderator: Jun 20, 2011
  6. Jun 20, 2011 #5
    When you get the answer, make sure you check it by taking the derivative - it's really cool how it works out! :)
  7. Jun 20, 2011 #6
    do the subst:
    z(x) = y(x) - x
    Then the variables in the new equation can be separated.
  8. Jun 21, 2011 #7
    I'm still unsure.

    I tried the following, I took the substition from HallsofIvy

    [tex] y'=z^{2}-1[/tex]

    Shall I just integrate both sides now? I think this doesn't work, as I still have y and x after the integration on one side.

    And I wonder about the following:

    I have [tex]y'=z^{2}-1 [/tex] but when I say I substitute [tex] x-y=z[/tex] and insert it in my given equation I have [tex] y'=z^{2}[/tex] :mad:
  9. Jun 21, 2011 #8

    Char. Limit

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    I think you misunderstood. The equation HallsofIvy got to is

    [tex]z' = z^2 - 1[/tex]

    Not y'.
  10. Jun 21, 2011 #9
    You performed the substitution incorrectly. If z = y - x, take d/dx of both sides: dz/dx = dy/dx - 1. Since dy/dx = (y - x)^2 = z^2, we have dz/dx = z^2 - 1.

    Now, separate variables and use partial fractions. To separate variables, integrate both sides of f(z) dz = g(x) dx. :)
  11. Jun 21, 2011 #10
    Thank you, just to make sure I didn't do any mistakes thus far. I have to do the partial fraction with the following:

    [tex] x= \int \frac1 {z^{2}-1} dz[/tex]
  12. Jun 21, 2011 #11
    Yes. For simplicity's sake, I'd put your constant of integration (+ C) on the x side.
  13. Jun 21, 2011 #12
    Wow, it's really embarrassing right now.

    My partial fraction looks like:

    [tex] \frac 1 {z^{2}-1} = \frac A {z-1} + \frac B {z+1} [/tex]
    A= \frac 1 2[/tex]
    B= -\frac1 2


    Now I tried to integrate


    x=\frac1 2 \int{\frac 1 {z-1}}dz - \frac 1 2 \int{\frac 1 {z+1}}dz [/tex]


    [tex] 2x+2c = ln(z-1)-ln(z+1)[/tex]

    but the z gets lost when I *exp on both sides. Where's my mistake. It's so frustrating :cry:
  14. Jun 21, 2011 #13
    Don't fret. Recall that ln(a) + ln(b) = ln(ab) and ln(a) - ln(b) = ln(a/b).

    Also, when you take the exponential of 2x + 2C, let A = e^(2C) for simplicity. Your left side should be Ae^(2x).
  15. Jun 21, 2011 #14
    Thanks for the help and patience :biggrin:

    my solution looks like this:

    I used the ln calculation rule:

    whereas, as you said, my A=exp(2c)

    [tex] z=\frac {1-A-e^{2x}} {-1+A+e^{2x}} [/tex]

    the only step I have to do now is to resubstitute z with [tex] z= x-y [/tex]

    and solve the equation, bringing y on one side, which would look like this:

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

    is this correct (and the final answer) ?
  16. Jun 21, 2011 #15
    No, but close. For any base k > 0, k^(a + b) does not equal k^a + k^b; it equals (k^a)(k^b).

    Also, we set z = y - x, not z = x - y.
  17. Jun 21, 2011 #16
    I should really try to concentrate some more :smile:

    Another attempt, A=exp(2c) like before

    final equation looks like:

    [tex] y =\frac{-1-Ae^{2x}} {-1+Ae^{2x}}+x[/tex]
  18. Jun 21, 2011 #17
    Yes! Well done. One more thing: you could decimate the abundance of minus signs (*shudder*) by multiplying with -1/-1 to get
    [tex]y = \frac{1 + Ae^{2x}}{1 - Ae^{2x}} + x[/tex]
    as your final answer.

    As an exercise, you should find y' and see that it does indeed equal (y - x)2. See post #5 :)
  19. Jun 21, 2011 #18
    Thank you very much :smile:
    I'll try that, I tried it for the specific A=0 because that worked quite fast, but I'll do it generally and see how it works. I really need some practice with differential equations.
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