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Math Help - Second order differential equation

  1. #1
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    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.
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  2. #2
    Senior Member Peritus's Avatar
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    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:

    <br />
\begin{gathered}<br />
  v = y^{ - 2}  \hfill \\<br />
  v' =  - 2y^{ - 3} y' \hfill \\ <br />
\end{gathered} <br />

    thus:

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


    now substitute the resulting expressions in the DE:

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

    multiply the equationn by: <br />
v^{{\raise0.7ex\hbox{$3$} \!\mathord{\left/<br />
 {\vphantom {3 2}}\right.\kern-\nulldelimiterspace}<br />
\!\lower0.7ex\hbox{$2$}}} <br />
    and you'll get a regular ODE...
    Last edited by Peritus; January 27th 2008 at 02:54 AM.
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  3. #3
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    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 v = \frac{dy}{dx}, it follows that \frac{dv}{dx} = \frac{d^2 y}{dx^2}.

    Therefore 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 \frac{dy}{dx} = v(x) for y = y(x).

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

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

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

    Sub into DE:

    -\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}

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

    Solve for v = v(x) using the integrating factor method. Then y^2 = \frac{1}{v(x)}.
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  4. #4
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    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?
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  5. #5
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    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 \frac{dv}{dx} - \frac{2}{x} v = -1, use the integrating factor method.

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

    It follows that v = x + C x^2.

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

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

    Then y = \frac{1}{2} x^2 + A x^3 + B.
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