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Find a general solution of the differential equation

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Find a general solution of the differential equation.

    [tex](x+y)y' = x-y[/tex]


    2. Relevant equations

    [tex]v=x+y[/tex]
    [tex]y=v-x[/tex]
    [tex]y'=v'-1[/tex]

    3. The attempt at a solution
    So if I plug this back into the original equation;
    [tex] v(v'-1)= x-y [/tex]
    How do I convert [tex]v=x+y[/tex] into [tex]x-y[/tex] so that I have only v's and x's? Do I have to make another substitution with x-y?
     
  2. jcsd
  3. Feb 19, 2012 #2
    Re: Ode

    Hint:

    Make the subst:

    [tex]
    z = \frac{y}{x}
    [/tex]
    because your ode is a homogeneous equation:
    [tex]
    y' = \frac{x - y}{x + y} = \frac{\frac{x - y}{x}}{\frac{x + y}{x}} = \frac{1 - y/x}{1 + y/x}
    [/tex]

    Treat z as a function of x. Exprss y in terms of x and z. Find the derivative w.r.t. x by using the chain rule. What do you get?
     
  4. Feb 19, 2012 #3
    Re: Ode

    So it would be;

    [tex]y=zx[/tex]
    [tex]y'= z+xz'[/tex]
    [tex] z+xz' =\frac {1-z}{1+z} [/tex]

    How do I isolate the z's and x's in order to integrate?
     
  5. Feb 19, 2012 #4
    Re: Ode

    simplify:
    [tex]
    \frac{1 - z}{1 + z} - z
    [/tex]
    You will get an ODE with separated variables x and z.
     
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