Separable Equations

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March 6th 2011, 02:59 PM
Naples

Separable Equations

1. Solve
$dP/dt = P(a-bP)$ where a and b are constants
$dP/[P(a-bP)] = dt$
rewrote the integral as
$-dP/[P(bP-a)] = dt$
then used this list of integrals [HTML]http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions[/HTML]
$1/a[ln((bP-a)/P)] = t + C$
multiply both sides by a and raise everything to e
$(bP-a)/P = Ce^a^t$

I'd appreciate it if someone could check this for me and tell me what I might have done wrong.

2. Find a solution that passes through the points (0, 0) and (0, 1)
$x dy/dx = y^2 - y$
$1/ (y^2 - y) = 1/x dx$
used partial fractions on left side to get
$[-1/y + 1/(y-1)] dy = 1/x dx$
$-ln y + ln(y - 1) = ln x + C$
$ln[(y-1)/y] = ln x + C$
$(y-1)/y = Cx$

I'm pretty sure I did something wrong, most likely with the partial fractions b/c if you put the given points into my answer it yields 0=0 and DNE so I'd appreciate if someone could tell me what I did wrong.

Thanks in advance.
March 6th 2011, 03:32 PM
zzzoak

Could you show the partial fractions of the expression

$<br /> \displaystyle<br /> \frac{1}{P(bP-a)}<br />$

and then integration here ?
March 6th 2011, 03:39 PM
topsquark

Quote:

Originally Posted by Naples

$-dP/[P(bP-a)] = dt$
$1/a[ln((bP-a)/P)] = t + C$
multiply both sides by a and raise everything to e
$(bP-a)/P = Ce^a^t$

The integral is written wrong. It should be:
$(1/a)[ln(P/(bP-a))] = t + C$

Also, you are raising the arbitrary constant to the e-th power, which is fine. However you have left it as an arbitrary constant in your final answer. (ie. you are calling it "C.") Please note that this new C you are talking about can never be negative, so it is not quite an arbitrary constant. I would suggest stating your solution and adding the comment that C > 0.

-Dan
March 6th 2011, 04:34 PM
topsquark

Quote:

Originally Posted by Naples

2. Find a solution that passes through the points (0, 0) and (0, 1)
$x dy/dx = y^2 - y$
$1/ (y^2 - y) = 1/x dx$
used partial fractions on left side to get
$[-1/y + 1/(y-1)] dy = 1/x dx$
$-ln y + ln(y - 1) = ln x + C$
$ln[(y-1)/y] = ln x + C$
$(y-1)/y = Cx$

For starters there is no way we can specify a solution of a first order ODE to pass through two arbitrary points. So the problem is automatically over-constrained from the start.

I did this problem as a Bernoulli equation and got a different solution.
$\displaystyle y(x) = \frac{1}{Ax + 1}$

I did it your way and got your solution. I am not certain why the two forms do not match up. Anyway, my solution can be passed through the point (0, 1), but not (0, 0).

-Dan

Wait. They do match up, but they aren't quite in the same form. Both forms can pass through (0, 1), but not (0, 0).