Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Solution for these Differential equations

  1. Oct 30, 2005 #1
    Hi everyone. I'm trying to solve these differential equations but I could not crack a single nut. I seem to have lost my memory on solving differential equations :bugeye:. Please help me refresh it by providing useful hints.
    I'm unable to separate the variables in the following. Perhaps I'm missing out on something important.

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

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

    3] [tex]x^2\frac{dy}{dx} - 2xy = \frac{1}{x}[/tex]
     
  2. jcsd
  3. Oct 30, 2005 #2

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    First one is a Riccati. Make the change of variables usually done for such equations and see what happens.
     
  4. Oct 30, 2005 #3

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    For the second one, can you simplfy the derivative term? After a well chosen substitution, the differential equation becomes linear.

    For the third one, the equation is linear first order and there is a general method available.
     
  5. Oct 31, 2005 #4
    Thank you so much for the help.

    Well, I was able to solve the third one!

    Bringing the equation in the general form:
    [tex]\frac{dy}{dx} + P(x)y = Q(x)[/tex]

    [tex]\frac{dy}{dx} - \frac{2}{x} y = \frac{1}{x^3}[/tex]

    Setting [tex]y = u(x)v(x)[/tex]

    So,
    [tex]\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}[/tex]

    [tex]u\left(\frac{dv}{dx} - \frac{2}{x}v\right) + v\frac{du}{dx} = \frac{1}{x^3}[/tex].....(1)

    Solving for v:
    [tex]\frac{dv}{dx} - \frac{2}{x} v = 0[/tex]

    On solving:
    [tex]\ln v = \ln x^2[/tex]
    [tex]v = x^2[/tex]

    Again on substituting for v in (1):
    [tex]u = -\frac{1}{4x^4} + C [/tex]

    General formula:
    [tex]y = v(x)\int \frac{Q(x)}{v(x)} dx + (C)v(x)[/tex]

    [tex]y = -\frac{1}{4x^2} + C{x}^2[/tex]
    Hope I'm right.
    Sorry I could not find any suitable substitution for the second one :frown: . Please help!
     
    Last edited: Oct 31, 2005
  6. Oct 31, 2005 #5
    Ricatti? I haven't studied any differential equation like that. How do you solve such equations.
     
  7. Oct 31, 2005 #6

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    First place it into standard form:

    [tex]y^{'}+Q(x)y+R(x)y^2=P(x)[/tex]

    Now, make the transformation:

    [tex]y=\frac{u^{'}}{Ru}[/tex]

    Can you now substitute this into the ODE? I'll start it for you:

    [tex]y^{'}=\frac{Ruu^{''}-u^{'}(Ru^{'}+uR^{'})}{(Ru)^2}[/tex]

    right?

    Make the other ones to get:

    [tex]\frac{Ruu^{''}-u^{'}(Ru^{'}+uR^{'})}{(Ru)^2}+\frac{Qu^{'}}{Ru}+R\left(\frac{u^{'}}{Ru}\right)^2=P[/tex]

    Now simplify and obtain a second order in u. Solve, convert back to y, and I want a plot.

    Edit: Suppose that last one looks a bit intimidating. That's just the general expression though. For your equation a lot of stuff just drops out leaving a simple second order to solve. Try it.
     
    Last edited: Oct 31, 2005
  8. Oct 31, 2005 #7

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    For the second one, focus on the term [tex] 2 y \frac{dy}{dx} [/tex], can you write this as something more convenient? Hint: notice that the only other y term you have is [tex] y^2 [/tex].
     
  9. Nov 1, 2005 #8
    I had tried it:

    Set,
    [tex]u = y^2[/tex]

    So,
    [tex]\frac{du}{dx} = 2y\left(\frac{dy}{dx}\right)[/tex]

    So the eqaution becomes,
    [tex]\frac{du}{dx} + u = \frac{x}{2} e^{-x}[/tex]

    But, I'm still unable to separate the variables. Should I adopt a different method?
     
  10. Nov 1, 2005 #9
    I don't think you can separate variables. You could try the integrating factor technique since you have a first order linear differential equation.
     
  11. Nov 1, 2005 #10

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Reshma, rearrange the equation to:

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

    or:

    [tex]\left(\frac{x}{2}e^{-x}-y^2\right)dx-2ydy=0[/tex]

    Now, we can make this exact right? You know, the partial of M with respect to y, partial of N with respect to x, do that arithmetic, get some function of x or y, then e to the integral of that function is the integrating factor right? You know this makes two plots now.
     
    Last edited: Nov 1, 2005
  12. Nov 1, 2005 #11

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    What Salty suggested is a nice way of proceeding, or as Benny said, you now have a linear first order equation in u and a general approach exists as I said before.
     
  13. Nov 2, 2005 #12
    Wow, thanks for your help, Saltydog and PhysicsMonkey. I got all the solutions!!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook