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Differential Equation

  1. Jan 11, 2013 #1
    1. The problem statement, all variables and given/known data

    y = 2xy' + y(y')2 ; y2 = c1(x + c1/4)

    2. Relevant equations

    So far I've gotten the second equation to be: y = (c1x + c12/4)1/2

    I was then going to take the derivative of that equation and plug them into the first equation after setting it to zero.

    Is that the right way to handle this particular equation?
     
  2. jcsd
  3. Jan 11, 2013 #2

    Mark44

    Staff: Mentor

    Re: Differential Eq.

    Your problem statement is incomplete - it doesn't say what you need to do. Are you supposed to show that the equation with y2 is a solution of the diff. equation?

    Also, your equation for y2 should be y2 = c1(x + c2/4)


    "after setting it to zero" - ??

    Your equation for y should be written as y = ±(c1(x + c2/4))1/2, since the original equation determines two values for y: one positive and one negative.

    Substitute your expressions for y and y' into the differential equation. If the solution is correct, you'll get an equation that is identically true.
     
  4. Jan 11, 2013 #3
    Re: Differential Eq.

    Mark44,

    You are correct, the problem asked to verify that the indicated function is a solution of the given differential equation.

    - My initial equation was correct. There is no c2 in the equation. (see attached document from text book)

    - "after setting it to zero" : I was referring to y = 2xy' + y(y')^2 --> 0 = 2xy' + y(y')^2 - y

    Thanks for your reply.
     

    Attached Files:

  5. Jan 11, 2013 #4

    Mark44

    Staff: Mentor

    Re: Differential Eq.

    That is possibly a typo in the book. The right side of that equation is the same as c1x + c2, where c2 = (1/4)c12.
    OK, that wasn't clear to me. You could work with the equation as-is, without moving y to the other side. It doesn't make much difference either way.
     
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