Question

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?

Answer 1

You want to solve the DE?

Answer 2

\(\color{#000000 }{ \displaystyle \frac{dx }{dt}=\frac{k+x}{x} }\)

(so x is a function of t)

Answer 3

You can do separation of variables.

Answer 4

\(\color{#000000 }{ \displaystyle \frac{dx }{dt}=\frac{k+x}{x} }\)

\(\color{#000000 }{ \displaystyle \frac{x}{k+x} \cdot \frac{dx }{dt}=1 }\)

\(\color{#000000 }{ \displaystyle \color{#ff0000}{\int}\frac{x}{k+x} \cdot \frac{dx }{\cancel{dt}}\color{#ff0000}{\cancel{dt}}=\color{#ff0000}{\int}1 \color{#ff0000}{dt}}\)

Answer 5

what I got was \[x-k \ln|x+k|+C=t\]

Answer 6

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!

Answer 7

now, you need to solve for x.

Answer 8

how do i go about pulling the variables out of ln|x=k|without turning the other x into an exponent of e?

Answer 9

you need to use the Albert .. blah (forgot the name) function W(x)

Answer 10

It is the inverse of xe^x..

Answer 11

this is a pretty hard equation to solve for x...

Answer 12

that is wolfram's result.
http://www.wolframalpha.com/input/?i=x-k+ln%28x%2Bk%29%2BC%3Dt++solve+for+x