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Thread: First Order Differential Equation

  1. #1
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    First Order Differential Equation

    I have solved a differential equation ($\displaystyle dx/dy-2xy=x$ if y(1)=0) and obtained this:
    $\displaystyle y=C e^(x^2)-0.5$



    I think i've got it right.. but ain't sure...
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by MissWonder View Post
    I have solved a differential equation ($\displaystyle dx/dy-2xy=x$ if y(1)=0) and obtained this:
    $\displaystyle y=C e^(x^2)-0.5$



    I think i've got it right.. but ain't sure...
    $\displaystyle x'=x+2xy$ Separate and integrate:

    $\displaystyle \int\frac{x'}{x}\,dy=\int(1+2y)\,dy \implies$ $\displaystyle \ln(x)=y+y^2+c_0 \implies \ln(x)=y^2+y+\frac{1}{4}+c_1$ $\displaystyle \implies \ln(x)=\left(y+\frac{1}{2}\right)^2+c_1\implies y=\sqrt{\ln(x)-c_1}-\frac{1}{2}$

    At $\displaystyle (1,0)$, we have $\displaystyle \sqrt{0-c_1}-\frac{1}{2}=0\implies c_1=-\frac{1}{4}$

    So $\displaystyle \boxed{y=\sqrt{\ln(x)+\frac{1}{4}}-\frac{1}{2}}$
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  3. #3
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    Quote Originally Posted by MissWonder View Post
    I have solved a differential equation ($\displaystyle dx/dy-2xy=x$ if y(1)=0) and obtained this:
    $\displaystyle y=C e^(x^2)-0.5$



    I think i've got it right.. but ain't sure...
    DE will be reduced to
    $\displaystyle \frac {dx}{x}=(1+2y)dy$
    $\displaystyle integrating\ gives \quad ln x=y+y^2+c$
    $\displaystyle y(1)=0 \ gives \quad ln 1=0+0+c \quad or \quad c=0$
    $\displaystyle \therefore \boxed{ ln x=y+y^2} \quad or\quad \boxed {y^2+y-ln x=0}$
    Last edited by ramiee2010; Oct 16th 2009 at 08:55 AM.
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  4. #4
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    Hello, MissWonder!

    Is that derivative inverted?


    I have solved a differential equation: . $\displaystyle \frac{dx}{dy}-2xy\:=\:x$ .for $\displaystyle y(1)=0$
    and obtained this: .$\displaystyle y\:=\:Ce^{x^2}-\tfrac{1}{2}$

    I think i've got it right.. but ain't sure...
    If it is really $\displaystyle \frac{dx}{dy}$

    . . then we have: . $\displaystyle \frac{dx}{dy} \;=\;2xy + x \;=\;x(2y+1)$

    Separate variables: .$\displaystyle (2y+1)dy \:=\:\frac{dx}{x}$

    Integrate: . $\displaystyle y^2 + y \;=\;\ln x + C$

    Since $\displaystyle y(1)=0$, we have: .$\displaystyle 0^2 + 0 \;=\;\ln(1) + C \quad\Rightarrow\quad C = 0$


    . . Therefore: .$\displaystyle y^2 + y \:=\:\ln x$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    If the equation is: . $\displaystyle {\color{blue}\frac{dy}{dx}}- 2xy \:=\:x$
    . . we have: . $\displaystyle \frac{dy}{dx}\:=\:2xy + x \quad\Rightarrow\quad \frac{dy}{dx} \:=\:x(2y+1)\quad\Rightarrow\quad \frac{dy}{2y+1} \:=\:x\,dx$


    Integrate: . $\displaystyle \tfrac{1}{2}\ln(2y+1) \:=\:\tfrac{1}{2}x^2 + C \quad\Rightarrow\quad \ln(2y+1) \:=\:x^2 + C $

    Then: . $\displaystyle 2y + 1 \:=\:e^{x^2+C} \;=\;e^{x^2}\!\cdot e^C \:=\:Ce^{x^2} $

    . . . . . $\displaystyle 2y \:=\:Ce^{x^2} - 1 \quad\Rightarrow\quad y \:=\:Ce^{x^2} - \frac{1}{2}$

    Since $\displaystyle y(1) = 0$, we have: .$\displaystyle 0 \:=\:C(e^1) -\frac{1}{2} \quad\Rightarrow\quad C \:=\:\frac{1}{2e}$


    Hence: .$\displaystyle y \;=\;\frac{1}{2e}e^{x^2} - \frac{1}{2} \;=\;\frac{1}{2}e^{x^2-1} - \frac{1}{2}$


    . . Therefore: .$\displaystyle y \;=\;\frac{1}{2}\left(e^{x^2-1} - 1\right) $

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