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Integration of a composite function

  1. Dec 10, 2013 #1
    y=f(x),K=F(y),
    and dF(y)/dy=y^2/y', (1)
    then
    dF(x)=y^2*dx;
    so, F(x)=int(y^2*dx)=int((f(x)^2)*dx);
    then we obtain,
    F(x)=(f(x))^3/(3*f'(x))+C;
    substitution of y=f(x) into F(x), we get, F(y)=y^3/(3*y')+C; (2)

    using the result above(eq.(2)), the dF(y)/dy can be computed as follow:

    dF(y)/dy=y^2/y' - (y^3*y'')/(3*(y')^3) (3)

    Probelm: Eq.(3) is not equal to the eq.(1).

    I don't know which step is not correct.
     
  2. jcsd
  3. Dec 10, 2013 #2

    ShayanJ

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    Gold Member

    I have trouble figuring out which part of your calculation is correct!
     
  4. Dec 10, 2013 #3

    HallsofIvy

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    What reason do you have for introducing "K" when it is not used again?

    I don't believe this is possible. Do you mean y' to be the dervative of y with respect to x?
    The derivative of F with respect to y has nothing to do with y's dependence of x.

    If you meant dF/dx, then [itex]dF/dx= (dF/dy)(dy/dx)[/itex]. If you meant that [itex]F(y)= y^3/3[/itex], then [itex]dF/dx= y^2[/itex] times y', not divided by it.

    and this doesn't follow from what you wrote before. I don't know where you got the [itex]y^2[/itex] but, again, if you meant that [itex]F(y)= y^3/3[/itex] then [itex]dF= y^3 y'(x)dx[/itex].

     
    Last edited: Dec 10, 2013
  5. Dec 10, 2013 #4
    Yes, you are correct that K is not used, we can get rid of it.


    Yes, y' means the derivative of y with respect ot x.

    So, you meant that we should consider y' as constant when computing the derivative of F with respect to y? I need more information to understand this.

    At the beginning, we have dF(y)/dy=y^2/y', then dF(y)/dy=y^2/(dy/dx),
    eliminate the dy on the two side of the equation, we obtain dF(y)=y^2*dx,
    substitution of y=f(x) into it, dF(f(x))=dF(x)=y^2*dx
     
  6. Dec 10, 2013 #5

    pwsnafu

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    Suppose that ##F(y) = y^3 / 3## and ##y## is a function of x. Then
    ##\frac{dF}{dy} = y^2##
    ##\frac{dF}{dx} = y^2 \frac{dy}{dx}##

    How? Where did this come from?

    You are using ##\frac{1}{dy/dx} = \frac{dx}{dy}## which requires y to be an invertible function.
     
  7. Dec 11, 2013 #6
    pwsnafu, thank you for your reply.
    dF(y)/dy=y^2/y'-----this equation is what I want to construct, it can be considered as a known condition, I want to get F(y).
     
    Last edited: Dec 11, 2013
  8. Dec 11, 2013 #7
    I find the problem.
    given that, [itex]\frac{dF(y)}{dy}=\frac{y^{2}}{y'}[/itex]
    then,
    [itex]\frac{dF(x)}{dx}=\frac{dF(y)}{dy}.\frac{dy}{dx}=\frac{y^{2}}{y'}.y'=y^{2}[/itex],
    so,
    [itex]F(x)=\int y^{2}.dx=\int f^{2}(x).dx[/itex]

    I made a mistake in the integration of F(x)
    i.e. [itex]F(x)≠\frac{f^{3}(x)}{3.f'(x)}+C[/itex]

    But, what should be the answer for F(x) from which I can obtain F(y)? should the specific expression for f(x) be given?
     
    Last edited: Dec 11, 2013
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