# Differntial Equation Trouble With Integral

• Feb 2nd 2009, 05:56 PM
zangestu888
Differntial Equation Trouble With Integral
the differintial equation given is:

$\displaystyle dx/dt=k(a-x)[sqrt(b-x)]$(a) I did part a finding the x as a function of t when a=b

part (B) is what am having trouble with it states;

If a>b, find t as a function of x. { hint use substitution u=(sqrt(b-x))

i tried the substituon i cant seem to simpify my work after sub, is for the left side

-1/2 integral_ sqrt(b-x)du/(a-x)(u^(1/2))
• Feb 2nd 2009, 08:14 PM
Jhevon
Quote:

Originally Posted by zangestu888
the differintial equation given is:

$\displaystyle dx/dt=k(a-x)[sqrt(b-x)]$(a) I did part a finding the x as a function of t when a=b

part (B) is what am having trouble with it states;

If a>b, find t as a function of x. { hint use substitution u=(sqrt(b-x))

i tried the substituon i cant seem to simpify my work after sub, is for the left side

-1/2 integral_ sqrt(b-x)du/(a-x)(u^(1/2))

ok, lets work on $\displaystyle \int \frac 1{(a - x) \sqrt{b - x}}~dx$

Let $\displaystyle u = \sqrt{b - x}$

$\displaystyle \Rightarrow u^2 = b - x \implies \boxed{u^2 - b + a = a - x}$

$\displaystyle \Rightarrow 2u~du = - dx$

$\displaystyle \Rightarrow - 2 u~du = dx$

So our integral becomes:

$\displaystyle -2 \int \frac u{(u^2 - b + a)u}~du = -2 \int \frac 1{u^2 - b + a}~du$

which you should be able to handle. (since a > b, -b + a > 0, so think "arctangent")