Question

dx/dt=k(a-x)

Answer 1

@DangerousJesse This is calculus, dx/dt is differentiation

Answer 2

dx/dt=k(a--x)

Answer 3

dx/dt = k(a-x)
dx = k(a-x) dt
int dx = int k(a-x) dt
x = kat - k int x dt
okay I don't know what to do next

Answer 4

@Adrian22 What does the double minus sign mean?

Answer 5

Solve the separable equation \[\frac{dx(t)}{ dt} = k(a-x(t))\]:
Divide both sides by \(k(a-x(t))\):
\[\frac{ dx(t)}{ {dt}}/{k(a-x(t))} = 1 \]Integrate both sides:
\[ \int\limits_1^{x(t)} \frac{1}{k(a-v)} dv = \int\limits_{c_1}^t 1 dv \]
Evaluate the integrals..

Answer 6

What is v?

Answer 7

Why set boundaries? I think this should be an indefinite integral instead of a definite integral

Answer 8

***
dx/dt = k(a-x)
dx = k(a-x) dt
int dx = int k(a-x) dt
***

ok, but you need the x's (and dx) on the same side. ditto for the t and dt
so divide both sides by (a-x) to get
\[ \int \frac{dx}{a-x} \ dx = \int k \ dt \]

Answer 9

integrating the right-hand side looks simple

for the right side, (**ignore typo, with double dx)
let u= a-x
du = d(a-x) = -dx so dx= -du
\[ \int - \frac{du}{u} = k \int dt \]

Answer 10

integral du/u is a standard form, and gives ln(u) + C

Answer 11

Yes your answer is correct, however it is assumed that k=\[\frac{ 2,303 }{ t }*\log(\frac{ b }{ a-x })\]