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Thread: differencial equation

  1. #1
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    differencial equation

    ive got solve the differential equation

    xyy' - 2(y+3)=0

    so i get







    The RHS is


    then i end up with


    1/2 (y-3ln|y+3|)=ln|x|+c

    which could go to

    $\displaystyle e^{y} = x^{2} + (y+3)^{3} + e^{c} $

    is this right, or have i re arranged it wrong at the end, and is this the best way to put the final answer?
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  2. #2
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    Hello, James!

    You did great . . . until the last step.


    then i end up with: .$\displaystyle \frac{1}{2}\left(y - 3\ln|y+3|\right) \:=\:\ln|x| + c$ . . . . Right!

    which could go to: .$\displaystyle e^y \:= \:x^2 + (y+3)^3 + e^c $ . . . . no

    We have: .$\displaystyle \frac{1}{2}\left(y - 3\ln|y+3|\right) \;=\;\ln|x| + C_1$

    Multiply by 2: .$\displaystyle y - 3\ln|y+3| \;=\;2\ln x + C_2$

    Then: .$\displaystyle y \;=\;2\ln x + 3\ln|y+3| + \ln(C) $

    . . . . . $\displaystyle y \;=\;\ln(x^2) + \ln(y+3)^3 + \ln(C)$

    . . . . . $\displaystyle y \;=\;\ln\left[Cx^2(y+3)^3\right]$


    Thus:. $\displaystyle e^y \;=\;Cx^2(y+3)^3$

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  3. #3
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    Look at Here
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, James!

    You did great . . . until the last step.



    We have: .$\displaystyle \frac{1}{2}\left(y - 3\ln|y+3|\right) \;=\;\ln|x| + C_1$

    Multiply by 2: .$\displaystyle y - 3\ln|y+3| \;=\;2\ln x + C_2$

    Then: .$\displaystyle y \;=\;2\ln x + 3\ln|y+3| + \ln(C) $

    . . . . . $\displaystyle y \;=\;\ln(x^2) + \ln(y+3)^3 + \ln(C)$

    . . . . . $\displaystyle y \;=\;\ln\left[Cx^2(y+3)^3\right]$


    Thus:. $\displaystyle e^y \;=\;Cx^2(y+3)^3$

    This should be
    $\displaystyle e^y = Cx^2|y + 3|^3$
    should it not?

    -Dan
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