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Math Help - Solving equation using Newton's Method

  1. #1
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    Solving equation using Newton's Method

    So I solved this complicated DE and got this equation and trying to solve for Vo(intial velocity). The instructions told us to use Newton's Method or a graphing program to solve for Vo(Matlab,Mathematica, Maple,etc.) Is there any other method to solve?

    Vo(-333.3)e^[-0.003*ln(98/0.03Vo)(-333.3)]-3266.67 ln(98/0.03V (-333.3)=10

    I have no idea where to start!
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  2. #2
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    Quote Originally Posted by latavee View Post
    So I solved this complicated DE and got this equation and trying to solve for Vo(intial velocity). The instructions told us to use Newton's Method or a graphing program to solve for Vo(Matlab,Mathematica, Maple,etc.) Is there any other method to solve?

    Vo(-333.3)e^[-0.003*ln(98/0.03Vo)(-333.3)]-3266.67 ln(98/0.03V (-333.3)=10

    I have no idea where to start!
    Mmm, difficult to understand what you mean. The equation is algebraic and not a DE, the brackets in the last logarithm don't match and there are two variables V and V0....

    Can you correct this first so I do not attempt to solve something wrong?

    Coomast
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  3. #3
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    sorry!

    Vo(-333.3)e^[-0.003*ln(98/0.03Vo)(-333.3)]-3266.67 ln(98/0.03Vo) (-333.3)=10

    Sorry! this was a very complicated one....yeah it started as a first order DE and was simplified in order to find Vo(initial velocity) We were told that it would be virtually impossible to solve this equation without using Newton's method or a graphing program...just wondered was there another way
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  4. #4
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    Quote Originally Posted by latavee View Post
    Vo(-333.3)e^[-0.003*ln(98/0.03Vo)(-333.3)]-3266.67 ln(98/0.03Vo) (-333.3)=10

    Sorry! this was a very complicated one....yeah it started as a first order DE and was simplified in order to find Vo(initial velocity) We were told that it would be virtually impossible to solve this equation without using Newton's method or a graphing program...just wondered was there another way
    Assuming that [imath]333.3=\frac{1000}{3}[/imath], and [imath]3266.67=\frac{9800}{3}[/imath], it can be shown that the equation can be rewritten as:

    V_0 \cdot e^{-V_0^2}=\frac{3}{9800} \cdot e^{\frac{9}{980000}}=A

    This equation can not be solved for [imath]V_0[/imath], however applying Newton-Raphson on the formula:

    f=x\cdot e^{-x^2}-A

    gives (x used instead of V0) as recursive equation:

    x_{n+1}=\frac{A \cdot e^{x_n^2}-2\cdot x_n^3}{1-2\cdot x_n^2}

    Start with [imath]x_0=0.1[/imath] and you will see that the solution is extremely fast found.

    Hope this helps,

    Coomast
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