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Math Help - First Order Differential Equation

  1. #1
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    First Order Differential Equation

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



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

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

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

    So \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 ( dx/dy-2xy=x if y(1)=0) and obtained this:
    y=C e^(x^2)-0.5



    I think i've got it right.. but ain't sure...
    DE will be reduced to
     \frac {dx}{x}=(1+2y)dy
     integrating\ gives \quad ln x=y+y^2+c
     y(1)=0 \ gives \quad ln 1=0+0+c \quad or \quad c=0
     \therefore \boxed{ ln x=y+y^2} \quad or\quad \boxed {y^2+y-ln x=0}
    Last edited by ramiee2010; October 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: . \frac{dx}{dy}-2xy\:=\:x .for y(1)=0
    and obtained this: . y\:=\:Ce^{x^2}-\tfrac{1}{2}

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

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

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

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

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


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


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


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


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

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

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

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


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


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

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