MHB Solve Differential Eq: xe^-1/(k+e^-1) for x, k, t

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The differential equation for the population of organisms, represented as dx/dt = xe^-t/(k+e^-t), can be solved by recognizing that e^-t is a variable dependent on time. By integrating the equation, the solution is derived as x(t) = 10e^(e^-t/(k+e^-t)t), with the initial condition x(0) = 10. The solution can also be expressed in a simplified form as x(t) = 10e^(t/(ke + 1)). This highlights the relationship between the population x, the constant k, and time t. The discussion emphasizes the importance of correctly identifying the variable in the exponential term for accurate solutions.
Shah 72
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The number of organisms in a population at time t is denoted by x. Treating x as a continuous variable, the differential equation satisfied by x and t is dx/dt= xe^-1/(k+e^-1), where k is a positive constant..
Given that x =10 when t=0 solve the differential equation, obtaining a relation between x, k and t.
Pls help. Iam not able to solve this
 
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Shah 72 said:
The number of organisms in a population at time t is denoted by x. Treating x as a continuous variable, the differential equation satisfied by x and t is dx/dt= xe^-1/(k+e^-1), where k is a positive constant..
Given that x =10 when t=0 solve the differential equation, obtaining a relation between x, k and t.
Pls help. Iam not able to solve this
I was able to solve it. I have done a silly mistake of writing e^-1. It's actually e^-t
 
Why in the world is a thread titled "differential equations" posted under Pre-Calculus?!

Shah 72, cogratulations on being able to do this.

For others who might be interested (and I really can't resist!):
Since $\frac{e^{-1}}{k+ e^{-1}}$ is a constant with respect to the variable, x, we can let $A= \frac{e^{-1}}{k+ e^{-1}}$ and write the equation as
$\frac{dx}{dt}= Ax$.

Then $\frac{dx}{x}= Adt$.

Integrating both sides, $ln(x)= At+ C$.

Taking the exponential of both sides, $x(t)= e^{At+ C}= C'e^{At}$
where $C'= e^C$ (but since C is an arbitrary constant so is C').

We are also told that x(0)= 10. $x(0)= C'e^0= C'= 10$ so
$x(t)= 10e^{At}= 10e^{\frac{e^{-1}t}{k+ e^{-1}}}$

We could also multiply both numerator and denominator of the exponent by $e$ and write the solution as
$x(t)= 10e^{\frac{t}{ke+ 1}}$
 
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