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Solving Simultaneous Equations

  • #1
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I need help solving these two equations simultaneously

[itex] y = c_1e^{r_1t_o}+c_2e^{r_2t_o}[/itex]

[itex]y' = c_1r_1e^{r_1t_o}+c_2r_2e^{r_2t_o}[/itex] My plan of solving these two equations is by substitution. By rearranging I obtain the following:

[itex]c_1 = [y-c_2e^{r_2t_o}]e^{-r_1t_o}[/itex]
[itex]c_1= \frac{[y' -c_2r_2e^{r_2t_o}]e^{-r_1t_o}}{r_1}[/itex]

Likewise,

[itex]c_2=[y-C_1e^{r_1t_o}]e^{-r_2t_o}[/itex]
[itex]c_2=\frac{[y'-c_1r_1e^{r_1e^r_1t_o}]e^{-r_2t_o}}{r_2}[/itex]

Don't know what to do from here. :confused:
 
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Answers and Replies

  • #2
64
1
Sorry, I don't see how these are two simultaneous equations. It's just the one equation you have for y and then you have its derivative with respect to t0 below it...
 
  • #3
1,291
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Marioeden[I said:
;3997417]Sorry, I don't see how these are two simultaneous equations. It's just the one equation you have for y and then you have its derivative with respect to t0 below it...
No, I am are trying to simultaneously solve from equations y & y' for c_1 & c_2. [/I]
 
  • #4
64
1
ah right, you want c1 and c2 in terms of y and y'

So like multiply the first equation by r2 and subtract them, then you'll get c1. Do the same thing but multiply the first equation by r1 instead and subtract them and you'll get c2 :)
 
  • #5
1,291
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Edit: Fixing Mistake

ah right, you want c1 and c2 in terms of y and y'

So like multiply the first equation by r2 and subtract them, then you'll get c1.
Okay so I get

[itex]c_1 = \frac{[y'-yr_2]}{r_2-r_1}e^{-r_1t_o}[/itex]

The book gets the same answer but with r_1-r_2 in the denominator. Any ideas why? As long as I subtract the first equation from the second, I will always get r_2-r_1.

P.s. I'll solve the other half after I grab some food.
 
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  • #6
64
1
You may wanna double check your algebra, the book is right...
 

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