differential equation problem

[hamzah]

At time t hours, the rate of decay of the mass of a radioactive substance is proportional to the mass x kg at that time.

 

(a) Write down the differential equation satisfied by x.

(b) Given that x = C when t = 0, show that x = Ce^[-(at)], where a is a positive constant.

 

I can't figure out part (a)...so I cant do part (b). So if somebody could help me with part (a) please.

 

thanks

[schlieffen]

(a) change of mass over change in time, (dx/dt) = ax (where a is the constant)

 

Now all you have to do is solve that DE and part (b) should work itself out

[bloodhound]

[math]

\[

\begin{array}{c}

\frac{{dx}}{{dt}} = - ax \\

\frac{1}{x}dx = - a{\rm{ }}dt \\

\ln x = - at + k \\

x = e^{ - at + k} \\

x = e^k e^{ - at} \\

x = Ce^{ - at} \\

\end{array}

\]

[/math]

at the second to last line. we know x=c when t=0 . pluggin the values in we get e^k = C

[Dave]

That's the basic decay equation that you'll get for a number of applications (radioactive decay, newton's cooling curve, capacitor discharge, etc).

 

Incidentally, the line where you say [math]\tfrac{1}{x} \, dx = -a \, dt[/math] is not technically correct. What you're actually doing is this:

 

[math]\frac{dx}{dt} = -ax \Rightarrow \int \frac{1}{x} \frac{dx}{dt} \, dt = \int -a \, dt[/math]

 

Then using the chain rule to 'cancel' the dt's from the left hand side integral.

[bloodhound]

yes dave. thanks for pointing that out. i was aware of that.

 

but treating dx/dt as a fraction is probably the better way to teach pple who are challenged in the art of calculus

[Dave]

Indeed, it just makes for better practice.