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Differential equation x^2*y'=y^2

  1. Apr 22, 2014 #1
    Hello,

    I have trouble solving the following differential equation.

    I am trying to learn how to solve that form of DEs.

    The DE is:

    x2*dy/dx = y2

    There are no initial-value problem, but the solution should be given such that y is defined for all x.

    The most important for me is to learn how to solve this type of DE.

    I tried solving it using the separable equation way.

    x2*dy/dx = y2 --> dy/y2 = dx/x2 and then integrate both sides to get -1/y = -1/x + C --> -y = -x + 1/C --> y = x - 1/C but apparently this doesn't solve the equation...

    Hope you guys can help me.

    Thanks in advance.

    Sincerely,
    Mr. Fest
     
  2. jcsd
  3. Apr 22, 2014 #2

    hilbert2

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    How do you solve the equation ##-1/y=-1/x+C## for ##y##? The answer is NOT ##-y=-x+1/C## !
     
  4. Apr 22, 2014 #3
    Hi,

    You are right. The time was around 2am over here when I wrote that. My brain must have been fried after 10 hours of math work.

    The answer is ofc:

    -1/y = -1/x + C --> -1 = y(-1/x + C) --> -1/(-1/x + C) = y --> -1/((-1+Cx)/x) --> -x/(-1 + Cx) = y

    And, the specific solution, that, y is defined for all real x gives us that C = 0, for if not, then for some x giving Cx = 1, y would not be defined.

    So, one could say that the initial-value problem is that y must be defined for all real x giving rise to C = 0 and ultimately y = -x/-1 --> y = x

    This satisfies: x^2*dy/dx = y^2 as this means that x^2*1 = x^2 and also, y is defined for all real x.

    If there is anything wrong with my answer, please correct me.
     
  5. Apr 25, 2014 #4

    vanhees71

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    Your implicit solution is correct, i.e., we indeed have
    [tex]\frac{1}{y}=\frac{1}{x}-C=\frac{1-C x}{x}.[/tex]
    This gives
    [tex]y(x)=\frac{x}{1-C x}.[/tex]
    Now it's easy to check that this indeed satisfies the given differential equation for any value of [itex]C[/itex].
     
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