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

[Shah 72]

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

 

[Shah 72]

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

 

[HOI]

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
$x(t)= 10e^{\frac{t}{ke+ 1}}$