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General solution formula of a differential equation

  1. Nov 8, 2009 #1
    Hello,

    The general solution of a differential equation for [tex]y'+P(x)y=G(x)[/tex] is

    [tex]y(x)=e^{-\int P(x)dx}[C+\int e^{\int P(x)dx}G(x)dx][/tex]

    for [tex]y'+xy=x[/tex]

    [tex]y(x)=e^{-\int xdx}[C+\int e^{\int xdx}xdx][/tex] i have

    [tex]y=Ce^{-\frac{x^2}{2}}+1[/tex]

    By the other solution

    [tex]\frac{dy}{dx}+xy=x \rightarrow \frac{dy}{dx}=x(1-y)\Rightarrow \frac{dy}{1-y}=xdx[/tex]

    integrating the both sides

    [tex]\int \frac{dy}{1-y}=\int xdx\Rightarrow \ln \left|1-y \right|=\frac{x^2}{2}+C

    \left|1-y \right|=Ce^{\frac{x^2}{2}}\Rightarrow y=1-Ce^{\frac{x^2}{2}}[/tex] or [tex]y=1+Ce^{\frac{x^2}{2}}[/tex]

    Why the two solutions are different and where is my mistake?Please help me.Thanks.
     
  2. jcsd
  3. Nov 8, 2009 #2
    when you integrate 1/(1-y), you forgot the negative. It should be - ln (1- y).
     
  4. Nov 8, 2009 #3
    ohh thanks l'Hopital:blushing:
     
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