How to solve this non lenear equation

  • #1
1,395
0
[tex]
v_0=100e^{\frac{-v_0}{100}}\\
\ln {v_0}=\ln{e^{\frac{-v_0}{100}}}^{100}\\
[/tex]
the answer is [tex]v_0=56.7[/tex]
how to find [tex]v_0[/tex]
??
 

Answers and Replies

  • #2
34,678
6,387
You can't solve this type of equation by algebraic means, but you can use approximation techniques to get an approximate solution, which v0 = 56.7 appears to be.
 
  • #3
Shooting Star
Homework Helper
1,977
4
[tex]
v_0=100e^{\frac{-v_0}{100}}\\
\ln {v_0}=\ln{e^{\frac{-v_0}{100}}}^{100}\\
[/tex]
the answer is [tex]v_0=56.7[/tex]
how to find [tex]v_0[/tex]
??
This is one way of doing it.

[tex]
v_0=100e^{\frac{-v_0}{100}}.
[/tex]


Putting [itex]
{\textstyle{{{\rm v}_{\rm 0} } \over {100}}} = x, x = e^{ - x} \Rightarrow x = 1 - x + x^2 /2,
[/itex] retaining up to the second order term after expanding [itex]e^{-x}.[/itex]

Now solve for x to get the approximate value. Justify why you are neglecting one value.
 

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