Differential Equations and exponential growth

**Yehia** · February 13th 2010, 01:30 PM

The volume of water in a solution during a chemical process varies with time and satisfies the equation: $dx/dt=-3x/(1 plus t)^2$ (sorry no plus sign on this laptop :P )

Anyway, initially at rest, t=0, x= 1000. Show that at time T the volume is given by: x=1000e^(-3t/(1 plus t))

And then prove that the volume of water tends to a limit as t tends to infinite and find that limit to the nearest litre.

Any help would be VERY appreciated coz I am so stumped. Thank you!!!!

**skeeter** · February 13th 2010, 01:42 PM

Originally Posted by Yehia

The volume of water in a solution during a chemical process varies with time and satisfies the equation: $dx/dt=-3x/(1 plus t)^2$ (sorry no plus sign on this laptop :P )

Anyway, initially at rest, t=0, x= 1000. Show that at time T the volume is given by: x=1000e^(-3t/(1 plus t))

And then prove that the volume of water tends to a limit as t tends to infinite and find that limit to the nearest litre.

Any help would be VERY appreciated coz I am so stumped. Thank you!!!!

$\frac{dx}{dt} = \frac{-3x}{(1+t)^2} <br />$

$\frac{dx}{x} = \frac{-3}{(1+t)^2} \, dt$

$\ln|x| = \frac{3}{1+t} + C$

$x = Ae^{\frac{3}{1+t}}$

$1000 = Ae^{3}$

$A = 1000e^{-3}$

$x = 1000e^{-3} \cdot e^{\frac{3}{1+t}}$

$x = 1000e^{-3 + \frac{3}{1+t}}<br />$

$x = 1000e^{\frac{-3t}{1+t}}$

**e^(i*pi)** · February 13th 2010, 01:44 PM

Originally Posted by Yehia

The volume of water in a solution during a chemical process varies with time and satisfies the equation: $dx/dt=-3x/(1 plus t)^2$ (sorry no plus sign on this laptop :P )

Anyway, initially at rest, t=0, x= 1000. Show that at time T the volume is given by: x=1000e^(-3t/(1 plus t))

And then prove that the volume of water tends to a limit as t tends to infinite and find that limit to the nearest litre.

Any help would be VERY appreciated coz I am so stumped. Thank you!!!!

$\frac{dx}{dt} = \frac{3x}{(1+t)^2}$

I'd go with separation of variables here

$\frac{dx}{3x} = \frac{dt}{(1+t)^2}$

Integrating both sides

$\frac{1}{3}\ln|x| = -\frac{1}{1+t} + C$

$ln|x| = -\frac{3}{1+t} + C$

$x = e^{-\frac{3}{1+t} + C} = C'e^{-\frac{3}{1+t}}$

where $C' = e^C$

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