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Y value for second derivative involving e^y

  1. Oct 28, 2011 #1
    1. The problem statement, all variables and given/known data
    If xy + 9e^y = 9e, find the value of y'' at the point where x = 0.

    2. Relevant equations
    product rule


    3. The attempt at a solution
    Okay so first I found the first derivative using implicit differentiation and I got:
    [itex]y'=\frac{-y}{x+9e^{y}}[/itex]

    then, I found the second derivative to be:
    [itex]y''=\frac{y(1+9e^{y}*y')}{(x+9e^{y})^{2}}-\frac{y'}{x+9e^{y}}[/itex]

    the website won't let me enter y' so I substituted it into the second derivative and then set x=0 and I got:
    [itex]y''=\frac{2y-y^{2}}{9e^{2y}}[/itex]

    I'm not quite sure if I'm doing this right so I'd like a second opinion. I only have one attempt left and I don't want to get it wrong!
     
  2. jcsd
  3. Oct 29, 2011 #2

    SammyS

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    Plug x = 0 into [itex]xy + 9e^y = 9e\,.[/itex] Then solve for y . Use that in your expression for y''.
     
  4. Oct 29, 2011 #3

    dynamicsolo

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    Since you are asked for the value of [itex]y''(0)[/itex], you don't really need to fuss too much with an expression for [itex]y''[/itex].

    First, are we agreed that y = 1 for x = 0 ? (I want to be careful, since you only have one attempt left on your computer-problem system.)

    I agree with your result for [itex]y'[/itex]. So we can say that [itex]y'(0) = -\frac{1}{9e} [/itex]. Does that sound all right?

    The first implicit differentiation of the original equation produced

    [tex] y + xy' + 9 \cdot e^{y} \cdot y' = 0 , [/tex]

    so a second implicit differentiation yields


    [tex] y' + ( y' + xy'' ) + ( 9 \cdot e^{y} \cdot y' \cdot y' + 9 \cdot e^{y} \cdot y'' ) = 0 \Rightarrow 2y' + xy'' + 9 \cdot e^{y} \cdot ( y' )^{2} + 9 \cdot e^{y} \cdot y'' = 0 . [/tex]

    You could now insert your results for [itex]y(0)[/itex] and [itex]y'(0)[/itex] , with x = 0 , then solve the resulting equation for [itex]y''[/itex].
     
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