Results 1 to 5 of 5

Thread: Differential equation

  1. #1
    Member roshanhero's Avatar
    Joined
    Aug 2008
    Posts
    184

    Differential equation

    How to solve the equation-
    (x^2+y^2)dx+2xydy=0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by roshanhero View Post
    How to solve the equation-
    (x^2+y^2)dx+2xydy=0
    I'd write it as $\displaystyle \frac{dy}{dx} = -\frac{1}{2} \left(\frac{x^2 + y^2}{xy} \right) = -\frac{1}{2} \left(\frac{x}{y} + \frac{y}{x}\right)$.

    Now make the usual substitution $\displaystyle \frac{y}{x} = v \Rightarrow y = xv$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    you can easily see that this is an exact ODE

    $\displaystyle \frac{\partial }
    {{\partial y}}\left( {x^2 + y^2 } \right) = \frac{\partial }
    {{\partial x}}2xy
    $

    thus we can find the solution as follows:

    $\displaystyle
    \psi (x,y) = \int {2xy} dy = xy^2 + f(x)$

    we know that:

    $\displaystyle
    \begin{gathered}
    \frac{{\partial \psi (x,y)}}
    {{\partial x}} = y^2 + f'(x) = x^2 + y^2 \hfill \\
    \Leftrightarrow f(x) = \frac{{x^3 }}
    {3} + k \hfill \\
    \end{gathered} $

    thus the implicit solution is:


    $\displaystyle \frac{{x^3 }}
    {3} + xy^2 = c$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member roshanhero's Avatar
    Joined
    Aug 2008
    Posts
    184
    Quote Originally Posted by mr fantastic View Post
    I'd write it as $\displaystyle \frac{dy}{dx} = -\frac{1}{2} \left(\frac{x^2 + y^2}{xy} \right) = -\frac{1}{2} \left(\frac{x}{y} + \frac{y}{x}\right)$.

    Now make the usual substitution $\displaystyle \frac{y}{x} = v \Rightarrow y = xv$.
    Thanks-
    I subsituted v=y/x and got the expression-
    2v/1+3v^2 dv=-dx/x.
    I tried to integrate the left side with respect to v, but I am just completely lost in this case.I wasnot able to integrate it
    So please suggest me the stage after this.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by roshanhero View Post
    Thanks-
    I subsituted v=y/x and got the expression-
    2v/1+3v^2 dv=-dx/x.
    I tried to integrate the left side with respect to v, but I am just completely lost in this case.I wasnot able to integrate it
    So please suggest me the stage after this.
    $\displaystyle \frac{dy}{dx} = -\frac{1}{2} \left(\frac{x}{y} + \frac{y}{x}\right) \Rightarrow x \frac{dv}{dx} = \frac{-3v^2 - 1}{2v}$

    (the algebra leading to this should be routine at this level. You do realise that if y = vx then dy/dx is v + x dv/dx, right?).

    This DE is seperable: $\displaystyle \frac{dx}{x} = \frac{-2v \, dv}{3v^2 + 1}$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Partial Differential Equation satisfy corresponding equation
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: May 16th 2011, 07:15 PM
  2. Replies: 4
    Last Post: May 8th 2011, 12:27 PM
  3. Replies: 1
    Last Post: Apr 11th 2011, 01:17 AM
  4. Partial differential equation-wave equation(2)
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: Sep 6th 2009, 08:54 AM
  5. Partial differential equation-wave equation - dimensional analysis
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Aug 28th 2009, 11:39 AM

Search Tags


/mathhelpforum @mathhelpforum