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Homework Help: Numerical Methods: Taylor Series for Diff Equation

  1. Mar 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation

    [tex]\frac{dy^2}{dx^2}=xy^2-2yy'+x^3+4[/tex]

    where

    [tex]y(1)=1[/tex]
    [tex]y'(1)=2[/tex]

    by means of the Taylor-series expansion to get the value of y at x=1.1. Use terms up to [tex]x^6[/tex] and [tex]\Delta x=0.1[/tex]


    3. The attempt at a solution

    I'm unsure as to how I should go about determining the coefficients for the Taylor expansion from the given equation. What I mean is that I don't know how to apply implicit differentiation techniques to higher order derivatives as in this case. Do we treat the derivatives in the equation as if they were variables? In other words, should I differentiate as follows (I'm using brackets to easier relate the the original term above to my subsequent derivative of said term):

    [tex]\frac{dy^3}{dx^3}=(y^2+2xyy')-(2y'+2yy'')+3x^2[/tex]

    ?

    Perhaps there's even an easier way that I'm missing?
    Any help will be greatly appreciated.

    phyz
     
    Last edited: Mar 8, 2010
  2. jcsd
  3. Mar 8, 2010 #2
    Anybody?
     
  4. Mar 8, 2010 #3

    vela

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    Is the idea to assume a solution in the form of a Taylor series and solve for the coefficients? If so, just write out the series, differentiate it to get y' and y'', plug them into the differential equation, and then solve for the coefficients by matching powers of (x-1).
     
  5. Mar 9, 2010 #4
    Hi Vela!

    Well, following the simple example in my textbook (why do they always give the straightforward examples?) we use a standard Taylor series expansion

    [tex]y(x)=y(x_0)+y'(x_0)(x-x_0)+\frac{y''(x_0)}{2!}(x-x_0)^2+\frac{y'''(x_0)}{3!}(x-x_0)^3+\frac{y^{(4)}(x_0)}{4!}(x-x_0)^4+\frac{y^{(5)}(x_0)}{5!}(x-x_0)^5+\frac{y^{(6)}(x_0)}{6!}(x-x_0)^6[/tex]

    where, if we let [tex]h=x-x_0[/tex] we'll end up with

    [tex]y(x)=y(x_0)+y'(x_0)h+\frac{y''(x_0)}{2}h^2+\frac{y'''(x_0)}{6}h^3+\frac{y^{(4)}(x_0)}{24}h^4+\frac{y^{(5)}(x_0)}{120}h^5+\frac{y^{(6)}(x_0)}{720}h^6[/tex]

    after which the idea then is to determine the derivatives of y(x) in order to obtain values for the coefficients of h, i.e. we know that

    [tex]y(1)=1[/tex]
    [tex]y'(1)=2[/tex]

    so that

    [tex]\frac{d^2y}{dx^2}(1)=xy^2-2yy'+x^3+4=(1)(1)^2-2(1)(2)+(1)^3+4=2[/tex]

    we must then obtain the equation for y''' from y'' and substitute the values for

    [tex]y(1)=1[/tex]
    [tex]y'(1)=2[/tex]
    [tex]y''(1)=2[/tex]

    into this equation, giving us the value for y''' which will help us determine the coefficient of [tex]h^3[/tex] in our Taylor series. So we continue until we have all the coefficients, finally substituting the value x=1.1 into the place of h to obtain the desired value of y.

    Does that make sense?

    My issue is that I do not know how to differentiate

    [tex]\frac{d^2y}{dx^2}(1)=xy^2-2yy'+x^3+4[/tex]

    ...

    phyz
     
  6. Mar 9, 2010 #5

    vela

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    That's almost correct. Remember you're differentiating with respect to x, so when you differentiate 2yy', you get

    [tex]\frac{d}{dx}(2yy') = 2(\frac{dy}{dx})y'+2y(\frac{dy'}{dx}) = 2y'^2+2yy''[/tex]
     
  7. Mar 10, 2010 #6
    First of all, thanks for your help, I truly appreciate it.

    If you don't mind checking one last time though (I'm learning something new here, please bear with me :smile:).

    We differentiated

    [tex]\frac{dy^2}{dx^2}=xy^2-2yy'+x^3+4[/tex]

    and got

    [tex]\frac{dy^3}{dx^3}=(y^2+2xyy')-(2y'^2+2yy'')+3x^2[/tex]
    [tex]\frac{dy^3}{dx^3}=y^2+2xyy'-2y'^2-2yy''+3x^2[/tex]

    Differentiating this yet again, should then give

    [tex]\frac{d^4y}{dx^4}=2y\frac{dy}{dx}+(2yy'+2x\frac{dy}{dx}y')-4\frac{dy'}{dx}-(2\frac{dy}{dx}y''+2y\frac{dy''}{dx})+6x[/tex]
    [tex]\frac{d^4y}{dx^4}=2yy'+(2yy'+2xy'y')-4y''-(2y'y''+2yy''')+6x[/tex]
    [tex]\frac{d^4y}{dx^4}=4yy'+2xy'^2-4y''-2y'y''-2yy'''+6x[/tex]

    Am I right? If this is correct, then I think I finally get it.

    Thanks again!
    phyz
     
  8. Mar 10, 2010 #7

    vela

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    Not quite. When you differentiated 2xyy', you should get three terms, not two, because there are three factors. You also forgot to use the chain rule when differentiating 2y'2.
     
  9. Mar 11, 2010 #8
    Cheers Vela...This thing is going to get really, really nasty towards [tex]\frac{y}{(6)}[/tex] isn't it?

    Guess I have to try again...:grumpy:
     
    Last edited: Mar 11, 2010
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