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DiffEq Question

  1. Dec 29, 2013 #1
    $$ xy'' - y' = 3x^{2} $$
    $$ y' = p $$
    $$ y'' = p' $$
    $$ xp' - p =3x^{2} $$
    $$ p' - \frac{1}{x}p = 3x $$
    after multiplying by the integrating factor we get..
    $$ \frac{1}{x}p' - \frac{1}{x^{2}}p =3 $$
    so $$ [\frac{1}{x}p]' = 3? $$

    I know that these two below are equal, but can someone please show HOW
    $$ \frac{1}{x}p' - \frac{1}{x^{2}}p $$ equals $$[\frac{1}{x}p]' $$

    Thank you!!
     
  2. jcsd
  3. Dec 29, 2013 #2

    Dick

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    You want to find d/dx of the product (1/x)*p(x). Use the product rule.
     
  4. Dec 29, 2013 #3

    LCKurtz

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    Use the product rule on that last equation.
     
  5. Dec 29, 2013 #4
    Awesome, I now understand that part. Now for any equation $$y' + a(x)y = b(x)$$ after multiplying the integrating factor, will my left side always be... $$ [e^{∫a(x) dx} * y]' ?$$
     
    Last edited: Dec 29, 2013
  6. Dec 29, 2013 #5

    LCKurtz

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    I think you can answer your own question. is$$
    \left(e^{∫a(x) dx} * y' + e^{∫a(x) dx} a(x)y\right)$$
    the derivative of $$
    e^{∫a(x) dx} * y$$
     
  7. Dec 29, 2013 #6
    yes, thank you LCKurtz!!
     
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