# Solving equation using Newton's Method

• Sep 27th 2009, 09:20 AM
latavee
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!
• Sep 27th 2009, 11:44 AM
Coomast
Quote:

Originally Posted by latavee
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
• Sep 27th 2009, 02:03 PM
latavee
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
• Sep 28th 2009, 10:31 AM
Coomast
Quote:

Originally Posted by latavee
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:

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

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

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

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

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

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

Hope this helps,

Coomast