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Math Help - Stuck on two differential equations

  1. #1
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    Stuck on two differential equations

    Hi,

    The first equation is $dy + ln x \, y \, dx = (4x +lny) \, dx$. It doesn't seem to matter how I move around the variables, I can't group the y terms with dy and the x terms with dx.

    The second equation is $x \frac{dy}{dx} = y^2 + y^2 \, lnx$. For this one I rearranged the terms to get $\frac{1}{y^2}dy = \frac{1}{x}dx + \frac{lnx}{x} \, dx$. Solving the integrals gave me $ -\frac{1}{y} = lnx + \frac{1}{2}(lnx)^2$. I don't see the problem with this but the answer key gave me something different. Have I made an error?

    Thanks.
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  2. #2
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    Re: Stuck on two differential equations

    for the 2nd equation you end up with

    $-\dfrac 1 y = \ln(x)+\dfrac 1 2 (\ln(x))^2 + C_1$

    $y = -\dfrac 1 {2\ln(x) + (\ln(x))^2 + C}$, $C=2 C_1$

    still looking at the first one

    could you post the original form of the first diff eq?
    Thanks from KevinShaughnessy
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  3. #3
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    Re: Stuck on two differential equations

    One obvious point is that you have forgotten the "constant of integration". But was the "something different" in the answer key?
    Thanks from KevinShaughnessy
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    Re: Stuck on two differential equations

    Quote Originally Posted by romsek View Post
    for the 2nd equation you end up with

    $-\dfrac 1 y = \ln(x)+\dfrac 1 2 (\ln(x))^2 + C_1$

    $y = -\dfrac 1 {2\ln(x) + (\ln(x))^2 + C}$, $C=2 C_1$

    still looking at the first one

    could you post the original form of the first diff eq?
    Hi Romsek, the equation I posted is the original form straight from the textbook.

    Quote Originally Posted by HallsofIvy View Post
    One obvious point is that you have forgotten the "constant of integration". But was the "something different" in the answer key?
    Hi HallsOfIvy, the answer key gives $y(1 + \ln x)^2 + cy + 2 = 0$.
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  5. #5
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    Re: Stuck on two differential equations

    Quote Originally Posted by KevinShaughnessy View Post
    Hi Romsek, the equation I posted is the original form straight from the textbook.



    Hi HallsOfIvy, the answer key gives $y(1 + \ln x)^2 + cy + 2 = 0$.
    Okay, you had $-\frac{1}{y}= ln(x)+ \frac{1}{2}(ln(x))^2+ C$ where "C" is the constant of integration. An obvious thing to do is to multiply through by 2y: $-2= 2yln(x)+ y(ln(x))^2+ 2Cy$

    That is the same as $y((ln(x))^2+ 2ln(x))+ 2Cy+ 2= 0$

    Complete the square! $y((ln(x))^2+ 2ln(x)+ 1- 1)+ 2Cy+ 2= y(ln(x)+ 1)^2- y+ 2Cy+ 2= 0$

    $y(ln(x)+ 1)^2+ (2C- 1)y+ 2= 0$

    Finally, since "C" is an undetermined constant, so is "2C- 1". Let c= 2C- 1 so that we have
    $y(ln(x)+ 1)^2+ cy+ 2= 0$
    Thanks from KevinShaughnessy
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