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Solving for C and Y(x) in differential equations

  1. Oct 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve the separable differential equation Subject to the initial condition y(0) = -10:

    7x-3y[tex]\sqrt{x^{2}+1}[/tex][tex]\frac{dy}{dx}[/tex] = 0


    2. Relevant equations
    Differential Equations


    3. The attempt at a solution
    I got up to the point where:
    -[tex]\frac{3}{2}[/tex]y[tex]^{2}[/tex]=-7[tex]\sqrt{x^{2}+1}[/tex]+C

    I solved for c=-143

    but when I plug it back in, I get a mess of an equation in solving for y(x) and each time I plug it in to check my answer, it comes up as wrong. Can anyone show me the mistake I am making? (My answer ends up being (-2/3(-7(x^2+1)^(1/2)-143))^(1/2) or some variation of that).

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 15, 2008 #2

    Mark44

    Staff: Mentor

    In your solution equation, solve for y so that you have y = {whatever} + C. That will make it easier to differentiate and to substitute in your DE.
     
  4. Oct 15, 2008 #3
    Ok, that sounds like a good idea. But I'm having trouble wrapping my head conceptually around the two square roots, since y^2 equals a term with a square root already in it. And does c go under the root? This may seem kind of trivial but I think this is why I've gotten this problem wrong 10 times already haha
     
  5. Oct 16, 2008 #4

    Mark44

    Staff: Mentor

    How about multiplying both sides of the equation you ended up with by -2/3? Then you would have y^2 = some stuff + C', where C' = (-2/3) * (-143).

    If you want to check your solution, you could take the derivative with respect to x implicitly (i.e., d/dx(y^2) = 2y dy/dx = d/dx(the stuff on the right side)). Then you could solve for dy/dx algebraically and substitute it back into the original equation. Might be a little simpler than a square root of a square root plus a constant. And yes, the constant would have to go under the outer radical. And you would have to consider positive and negative roots.
     
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