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Help me with Linear equation pls ~

  1. Feb 20, 2005 #1
    (a)Show that xy' + 2y = 3x has only one solution defined at x=0.
    Then Show that the initial value problem for this equation with initial condition y(0)= yo has a unique solution when yo = 0 and no solution when y0=/= 0 .

    (b) Show that xy'-2y=3x has an infinte number of solutions defined at x=0. then show that the initial value problem for this equation with initial condition y(0) =0 has an infinite number of solutions.
     
  2. jcsd
  3. Feb 21, 2005 #2

    saltydog

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    How about (a):

    Set it up in std. form for first order ODE, calc integrating factor, solve. You get:

    [tex]y(x)=x+\frac{y_0}{x^2}[/tex]

    Well, the only way for this to have a solution in the Reals for initial value problem is for [itex]y_0=0[/itex] in which case, solution is [itex]y=x[/itex]. However, if y(0)[itex]\neq 0[/itex], then no solution exists at x=0 since this is indeterminate.

    Is there a more rigorous way to say this?
     
  4. Mar 2, 2005 #3
    and (b) gives:

    (.....i'll spare the details but it's the usual integrating factor prob...)

    [tex] y = x(cx - 3) [/tex]

    where c is your integration constant.

    To satisfy the IC y(0) = 0, c can take on any value you want.
     
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