anonymous
  • anonymous
I'm doing differential equations, and the problem is about oil leaking from a pipe line. My equation is dx/dt=(k+x)/x where k is a constant and x is the amount of oil. Can anyone help me out?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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SolomonZelman
  • SolomonZelman
You want to solve the DE?
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle \frac{dx }{dt}=\frac{k+x}{x} }\) (so x is a function of t)
SolomonZelman
  • SolomonZelman
You can do separation of variables.

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SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle \frac{dx }{dt}=\frac{k+x}{x} }\) \(\color{#000000 }{ \displaystyle \frac{x}{k+x} \cdot \frac{dx }{dt}=1 }\) \(\color{#000000 }{ \displaystyle \color{#ff0000}{\int}\frac{x}{k+x} \cdot \frac{dx }{\cancel{dt}}\color{#ff0000}{\cancel{dt}}=\color{#ff0000}{\int}1 \color{#ff0000}{dt}}\)
anonymous
  • anonymous
what I got was \[x-k \ln|x+k|+C=t\]
SolomonZelman
  • SolomonZelman
u=k+x u-k=x integrand is: (u-k)/u du 1 - k(u)\(^{-1}\) du antideriv. u - kln(u) +C so you get; k+x - kln|x+k|+C=t and k dissolves with C. GOOD!
SolomonZelman
  • SolomonZelman
now, you need to solve for x.
anonymous
  • anonymous
how do i go about pulling the variables out of ln|x=k|without turning the other x into an exponent of e?
SolomonZelman
  • SolomonZelman
you need to use the Albert .. blah (forgot the name) function W(x)
SolomonZelman
  • SolomonZelman
It is the inverse of xe^x..
SolomonZelman
  • SolomonZelman
this is a pretty hard equation to solve for x...
SolomonZelman
  • SolomonZelman
that is wolfram's result. http://www.wolframalpha.com/input/?i=x-k+ln%28x%2Bk%29%2BC%3Dt++solve+for+x

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