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First-order equations causing some trouble (for newbie)

  1. Mar 17, 2012 #1

    Pzi

    User Avatar

    Hi.

    So I'm new to differential equations (it has been a month since I started attending this course).

    Code (Text):
    \[3{x^2} - y = y'\sqrt {{x^2} + 1} \]
    http://img641.imageshack.us/img641/9833/eqn7464.png [Broken]

    A linear equation, quite obviously... However I'm pretty much not able to proceed due to complicated integration.
    And I expected something reasonably elegant...

    ----------

    Code (Text):
    \[5y + {(y')^2} = x(x + y')\]
    http://img36.imageshack.us/img36/4629/eqn8105.png [Broken]

    I tried to solve for y' , which seems to be
    Code (Text):
    \[y' = \frac{{x \pm \sqrt {5{x^2} - 20y} }}{2}\]
    http://img99.imageshack.us/img99/3154/eqn6514.png [Broken]

    But once more I am not able to proceed...
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 18, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that's just not a good way to proceed. Not every differential equation can be immediately reduced to an integral.

    You are correct that this equation is linear which means it is relatively easy to find an "integrating factor".

    We can write the equation as [itex]y'\sqrt{x^2+ 1}+ y= 3x^2[/itex]. That is equivalent to [itex]y'+ (x^2+1)^{-1/2} 2x^2(x^2+1)^{-1/2}[/itex]

    Now we look for a function, [itex]\mu(t)[/itex], such that
    [tex]\mu(x)(y'+ (x^2+1)^{1/2}y)= \mu(x) y'+ \mu(x)(x^2+1)^{-1/2}y= \frac{d(\mu y)}{dx}[/tex]
    ([itex]\mu[/itex] is the "integrating factor".)

    Since, by the product rule, [itex]d(\mu(x)y)/dx= \mu y'+ \mu'y[/itex] we must have [itex]\mu'= (x^2+1)^{-1/2}/mu[/itex] which is a separable equation for [itex]\mu[/itex]:
    [tex]\frac{d\mu}{\mu}= (x^2+ 1)^{-1/2}dx[/tex]
     
  4. Mar 18, 2012 #3

    Pzi

    User Avatar

    Thanks for response!

    [tex]
    \begin{array}{l}
    \frac{{d\mu }}{\mu } = \frac{{dx}}{{\sqrt {{x^2} + 1} }}\\
    \ln \left| \mu \right| = \ln \left| {x + \sqrt {{x^2} + 1} } \right|\\
    \mu = x + \sqrt {{x^2} + 1}
    \end{array}
    [/tex]

    Which means

    [tex]
    \begin{array}{l}
    y' + \frac{y}{{\sqrt {{x^2} + 1} }} = \frac{{3{x^2}}}{{\sqrt {{x^2} + 1} }}\\
    y' \cdot (x + \sqrt {{x^2} + 1} ) + \frac{{y(x + \sqrt {{x^2} + 1} )}}{{\sqrt {{x^2} + 1} }} = \frac{{3{x^2}(x + \sqrt {{x^2} + 1} )}}{{\sqrt {{x^2} + 1} }}\\
    \left( {y \cdot (x + \sqrt {{x^2} + 1} )} \right)' = \frac{{3{x^3}}}{{\sqrt {{x^2} + 1} }} + 3{x^2}\\
    y \cdot (x + \sqrt {{x^2} + 1} ) = {x^3} + \int {\frac{{3{x^3}dx}}{{\sqrt {{x^2} + 1} }}}
    \end{array}
    [/tex]

    And I'd be good to go from here. Can you confirm?
     
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