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Initial-value problem

  1. Jan 10, 2016 #1
    Wondering why getting different values of "C" depending on how I solve the question. Not sure the values are different. Thanks.

    1. The problem statement, all variables and given/known data

    Solve the initial value problem [itex]cos(x)Ln(y) \frac{dy} {dx} =ysin(x) [/itex], y>0, y(0)=e2.

    2. Relevant equations
    N/A.

    3. The attempt at a solution

    [itex]∫ \frac{Ln(y)} {y} dy=∫ \frac{sin x} {cos x} dx [/itex]

    u=Ln(y)
    [itex]du=\frac{1} {y}[/itex] dy
    dy=y du​

    [itex]∫ \frac{u} {y} y du=∫ tan(x) dx [/itex]
    ## \frac{1} {2}u^2+C_1=Ln|sec(x)| + C_2 ##
    ##\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3 ## ,Since ## C=C_2-C_1 ##
    ## Ln(y)=± \sqrt{2 Ln|sec(x)|+C} ## Equation #1
    ## y=e^{± √(2Ln|sec(x)|+C)} ##
    --------------
    ## e^2=e^{+√(2Ln|sec(x)|+C)} ## Since y(0)=e2. Also, exponent cannot be -ve since LS.=RS.
    ## e^2=e^{+√(2(0)+C)} ##
    ##2=√C ##
    ## C=4 ##

    Therefore ## y=e^{± √(2Ln|sec(x)|+4)} ## <--- ANS #1

    However, if we try solving for C using eqn #1:

    ## \frac{1}{2}Ln^2(y)=Ln|sec(x)|+C ##
    ## \frac{1}{2}Ln^2(e^2)=Ln|sec(0)|+C ##
    ## \frac{1}{2}(2^2)Ln^2(e)=Ln(1)+C ##
    ## 2(1)=0+C ##
    Therefore, ## y=e^{± √(2Ln|sec(x)|+2)} ## <----ANS #2.

    However, obviously ANS #1 ≠ ANS #2. Can someone please help me understand why and what I did wrong?
    Thanks.
     
    Last edited: Jan 10, 2016
  2. jcsd
  3. Jan 10, 2016 #2

    Samy_A

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    You don't use eqn #1 in the second part, but the equation immediately above it:
    ##\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3 ##
    Notice that ##C_3## is not equal to your final ##C##.
     
  4. Jan 10, 2016 #3

    mfb

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    Staff: Mentor

    The two "C" here differ by a factor of 2.
    Approach 1 is correct, approach 2 gives you C3 = 1/2 C = 2 which leads to C=4 as well.
     
  5. Jan 10, 2016 #4
    I'm not sure I follow.

    To solve for the correct answer for "C" (which is C=4 according to the textbook), all I did was rearrange ## \frac{1}{2}Ln^2(y)=Ln|sec(x)|+C ## to solve for "y" and then subbing in the initial values x=0, and y=e^2 gives you ## e^2=e^{\sqrt{C}} ## from which you get C=4. But, when I sub in the initial values directly into ## \frac{1}{2}Ln^2(y)=Ln|sec(x)|+C ## and rearrange for C I get a different answer. It's unclear to me why I'm getting different answers because in both cases the variable C is the same. Solving for C two different ways when you're only rearranging the equation differently to find C should give you the same answer if you apply the rules of logarithms and exponents correctly. The fact that I'm not getting the same value of C means that I'm not understanding some basic concept of math or that I'm making a mistake when I'm rearranging and solving for C.
     
  6. Jan 10, 2016 #5

    ehild

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    You forgot to multiply C2-C1 by 2.
     
  7. Jan 11, 2016 #6

    Samy_A

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    (bolding mine)
    What I bolded is wrong.
    When you solve for ##C## using ## y=e^{± √(2Ln|sec(x)|+C)} ##, you find ##C=4##. That is correct.
    When you solve for ##C_3## using ##\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3 ##, you find ##C_3=2##. That is correct too.
    But, as mfb, ehild and myself have mentionned, the constants in these two equations are not equal. You confused yourself by renaming ##C_3## to ##C## when you tried to find the constant using ##\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3 ##.
    Let's rewrite you equations keeping ##C_3##:
    ##\frac{1} {2}Ln^2 (y)=Ln|sec(x)| + C_3 ##
    ## Ln(y)=± \sqrt{2 Ln|sec(x)|+2C_3} ##
    ## y=e^{± √(2Ln|sec(x)|+2C_3)} ##.

    You see? Your final ##C=2C_3##, that's why you find 4 using one equation, and 2 using the other.
     
    Last edited: Jan 11, 2016
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