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Calc 2 differential equations

  1. Apr 13, 2013 #1
    im in calculus 2 right now and we are doing differential equaitons right now. im confused as to why when i find the integrating factor I(x)=e^(∫p(x) and when i multiply both sides i get
    e^∫(p(x))[(dy/dx)+p(x)*y]=d(e^(p(x)dx)*y) how are they equal. i will give an example.

    (dy/dx)+y=x*e^(x) the integrating factor is e^(∫dx)=e^(x) then i multiply both sides by
    e^(x) which gives

    e^(x)*[(dy/dx)+y]=e^(x)*[x*e^(x)]

    which is equal to:

    d[e^(x)*y]/dx=e^(x)*x; basically my question is how do you get here? how does e^(x)*dy/dx just disappear after i multiply the integrating factor to both sides?

    please help with what happens after this step: e^(x)*[(dy/dx)+y]=e^(x)*[x*e^(x)]
     
  2. jcsd
  3. Apr 14, 2013 #2

    Simon Bridge

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    Lets see... your example is: ##\renewcommand{\dyx}[1]{\frac{d #1}{dx}}##

    $$\dyx{y}+y=xe^{x}$$ - integrating factor is: ##e^{\int dx}=e^x##
    multiplying both sides by the integrating factor gives you:

    $$e^x\left [ \dyx{y}+y \right ] = e^x\left [ xe^x \right ]$$ ... to from there to the end result - you must first multiply out the brackets.

    But you can see that the two expressions are the same by doing the differentiation on the LHS:

    $$\dyx{} [e^x y]$$

    This is how you choose the integrating factor - you have to notice that this relation will work.
    http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx
     
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