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Homework Help: Ode seperable eq.

  1. Feb 2, 2009 #1


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    I'm getting confused as to what to do with the C's.

    1. The problem statement, all variables and given/known data
    [tex]\frac{{dy}}{{dx}} + 2xy = 0[/tex]

    2. Relevant equations

    Back of Book: [tex]y\left( x \right) = C\exp \left( { - x^2 } \right)[/tex]

    3. The attempt at a solution
    \frac{{dy}}{{dx}} = - 2xy\,\,\,\, \Rightarrow \,\,\,\,\frac{{dy}}{y} = - 2x\,dx\,\,\,\, \Rightarrow \,\,\,\,\frac{1}{y}dy = - 2x\,dx \\
    \int_{}^{} {\frac{1}{y}dy} = \int_{}^{} { - 2x\,dx} \\
    \ln \left( y \right) = \frac{{ - 2x^2 }}{2} + C\,\,\,\, \Rightarrow \,\,\,\,\ln \left( y \right) = - x^2 + C\,\,\,\, \Rightarrow \,\,\,\, \\
    \exp \left( {\ln \left( {y + C} \right)} \right) = \exp \left( { - x^2 + C} \right)\,\,\,\, \Rightarrow \,\,\,\, \\
    y + C = \exp \left( { - x^2 + C} \right)\,\,\,\, \Rightarrow \,\,\,\,y = \exp \left( { - x^2 + C} \right) - C \\

    The C from the y part is not necessarily the same C as from the x part, is it? How did they end up with only 1 C, and how did they get their C out of the parenthesis to end up multiplying by the right side?

    Thanks in advance!
  2. jcsd
  3. Feb 2, 2009 #2
    When you integrate both sides, you get C1 on the y side & C2 on the x side. You can combine them, to make C2-C1=C leaving it on the x side.
  4. Feb 2, 2009 #3
    Also, I don't see where you've typed out your last question. Kind of confusing me as well.

    Maybe this will help. e^(x+y)=e^x e^y
  5. Feb 2, 2009 #4


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    Where did this [itex]C[/itex] come from?

    If you want to include a constant of integration for [tex]\int\frac{dy}{y}[/tex] You should have [tex]\int\frac{dy}{y}=\ln(y)+C\neq\ln(y+C)[/tex]

    Also, you should label your constants so that you don't confuse them with each other:


    [tex]\int -2x dx=-x^2+C_2[/tex]

    [tex]\implies \ln(y)+C_1=-x^2+C_2 \implies \ln(y)=-x^2 +(C_2-C_1)[/tex]

    but [itex](C_2-C_1)[/itex] is just going to be some constant, you may as well call [itex]C_3[/itex]

    [tex]\implies \ln(y)=-x^2+C_3 \implies y=e^{-x^2+C_3}=e^{C_3}e^{-x^2}[/itex]

    And then just call [itex]C\equiv e^{C_3}[/itex].
  6. Feb 3, 2009 #5


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    Thanks rocomath. I understand up to e^(-x^2) e^C . But how does that turn into Ce^(-x^2) ?
    For example, 2^3 * 2^4 doesn’t turn into 4*2^3.
    128 <> 32
  7. Feb 3, 2009 #6
    e is a positive constant, C is a constant, positive constant^constant = constant. e^C = A - better?
  8. Feb 3, 2009 #7


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    Thanks, gabba^2hey! C3 did the trick!
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