Results 1 to 5 of 5

Thread: Second order differential equation

  1. #1
    Member
    Joined
    Sep 2007
    Posts
    198

    Second order differential equation

    1. Find the general solution of each differential equation using the substitution given.
    x(d^2 y)/(d x^2) – 2 dy/dx + x = 0; dy/dx = v

    2. Use the substitution v = y^-2 to find the general solution of the differential equation
    dy/dx + y/x = x^2 y^3
    Find also y in terms of x, given that y = 1 at x = 1.


    Please help me to solve these.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    1. Find the general solution of each differential equation using the substitution given.
    x(d^2 y)/(d x^2) – 2 dy/dx + x = 0; dy/dx = v
    what exactly don't you understand?, it's stright forward, you substitute and get a first order ODE.

    2.

    now use the chain rule:

    $\displaystyle
    \begin{gathered}
    v = y^{ - 2} \hfill \\
    v' = - 2y^{ - 3} y' \hfill \\
    \end{gathered}
    $

    thus:

    $\displaystyle
    \begin{gathered}
    y = v^{ - {\raise0.7ex\hbox{$1$} \!\mathord{\left/
    {\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$2$}}} \hfill \\
    y' = - \frac{1}
    {2}y^3 v' \hfill \\
    \end{gathered}
    $


    now substitute the resulting expressions in the DE:

    $\displaystyle
    - \frac{1}
    {2}v^{ - {\raise0.7ex\hbox{$3$} \!\mathord{\left/
    {\vphantom {3 2}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$2$}}} v' + \frac{{v^{ - {\raise0.7ex\hbox{$1$} \!\mathord{\left/
    {\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$2$}}} }}
    {x} =
    $$\displaystyle
    x^2 v^{ - {\raise0.7ex\hbox{$3$} \!\mathord{\left/
    {\vphantom {3 2}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$2$}}}
    $

    multiply the equationn by: $\displaystyle
    v^{{\raise0.7ex\hbox{$3$} \!\mathord{\left/
    {\vphantom {3 2}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$2$}}}
    $
    and you'll get a regular ODE...
    Last edited by Peritus; Jan 27th 2008 at 02:54 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by geton View Post
    1. Find the general solution of each differential equation using the substitution given.
    x(d^2 y)/(d x^2) – 2 dy/dx + x = 0; dy/dx = v

    2. Use the substitution v = y^-2 to find the general solution of the differential equation
    dy/dx + y/x = x^2 y^3
    Find also y in terms of x, given that y = 1 at x = 1.


    Please help me to solve these.
    1. Since $\displaystyle v = \frac{dy}{dx}$, it follows that $\displaystyle \frac{dv}{dx} = \frac{d^2 y}{dx^2}$.

    Therefore $\displaystyle x \frac{dv}{dx} - 2 v + x = 0 \Rightarrow \frac{dv}{dx} - \frac{2}{x} v = -1 \, $.

    Use integrating factor method to solve for v = v(x).

    Then solve $\displaystyle \frac{dy}{dx} = v(x)$ for y = y(x).

    --------------------------------------------------------------------------

    2. $\displaystyle v = \frac{1}{y^2} \Rightarrow \frac{dv}{dy} = -\frac{2}{y^3}$.

    Using the chain rule: $\displaystyle \frac{dy}{dx} = \frac{dy}{dv} \times \frac{dv}{dx} = -\frac{y^3}{2} \times \frac{dv}{dx}$.

    Sub into DE:

    $\displaystyle -\frac{y^3}{2} \frac{dv}{dx} + \frac{y}{x} = x^2 y^3 \Rightarrow -\frac{y^2}{2} \frac{dv}{dx} + \frac{1}{x} = x^2 y^2 \Rightarrow - \frac{1}{2v} \frac{dv}{dx} + \frac{1}{x} = \frac{x^2}{v}$

    $\displaystyle \Rightarrow \frac{dv}{dx} - \frac{2}{x} v = -2x^2$.

    Solve for v = v(x) using the integrating factor method. Then $\displaystyle y^2 = \frac{1}{v(x)}$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2007
    Posts
    198
    For question 1, it may need complementary function in this form Ae^mx + Be^mx.

    And my book’s answer is
    Ax^3 + ˝ x^2 + B.

    How can I found complementary function by this equation?
    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 geton View Post
    For question 1, it may need complementary function in this form Ae^mx + Be^mx.

    And my book’s answer is
    Ax^3 + ˝ x^2 + B.

    How can I found complementary function by this equation?
    What's all this complementary function business?? As I've already said, to solve $\displaystyle \frac{dv}{dx} - \frac{2}{x} v = -1$, use the integrating factor method.

    The integrating factor is $\displaystyle e^{\int -\frac{2}{x} \, dx} = e^{-2 \ln |x|} = e^{\ln x^{-2}} = \frac{1}{x^2}$.

    It follows that $\displaystyle v = x + C x^2$.

    Then $\displaystyle \frac{dy}{dx} = x + C x^2 \Rightarrow y = \frac{1}{2} x^2 + \frac{C}{3} x^3 + B$.

    Since $\displaystyle \frac{C}{3}$ is also an arbitrary constant, replace it with A, say.

    Then $\displaystyle y = \frac{1}{2} x^2 + A x^3 + B$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Sep 21st 2011, 03:37 PM
  2. Please help with this First Order Differential Equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Jan 10th 2010, 09:28 AM
  3. Replies: 2
    Last Post: Nov 25th 2008, 09:29 PM
  4. First order differential equation
    Posted in the Calculus Forum
    Replies: 9
    Last Post: Apr 5th 2008, 03:08 PM
  5. second order differential equation help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 2nd 2008, 07:11 AM

Search Tags


/mathhelpforum @mathhelpforum