# Differential equation. seperate variables and solve using partial fractions!

## Homework Statement

seperate and solve using partial fractions

dx/dt=9-4x2, x(0)=0

## The Attempt at a Solution

rearranging gives dx/(9-4x2) = dt
factorising denominator in preperation for partial fractions becomes

dx/(3-2x)(3+2x) then A/(3-2x) + B/(3+2x) dx

so A(3+2x)+ B(3-2x) = 1

therefore 3A+3B=1 and 2Ax-2Bx=0 therefore A+B=1/3 and A-B=0 therefore A=1/6 and B=1/6

so integral becomes

(1/6)/(3-2x) + (1/6)/(3+2x) dx = 1/6*-1/2ln|3-2x| + 1/6*1/2ln|3+2x|

and RHS is just t+C

with x(0)=0 then C = 0

the Answer is given as x(t)=[3(1-e-12t)]/[2(1+e-12t)]

but i cant seem to manipulate my answer to get there...

your partial fractions look correct and then when you integrated it, it also looks correct. Just combine the logs and then raise both sides to e and then solve for x .

With c=0...
-1/12ln((3-2x)/(3+2x))=t
Ln((3-2x)/(3+2x))=-12t
(3-2x)/(3+2x)=e^-12t
... ?
How to gather x terms together? This Is where icant get to the answer given :(

multiply both side by (3+2x) then get all the x's on one side and then factor out the x

Got it!
A case of staring at it for too long and missing the basic method. Just needed reassuring I was doing it right.
Thanks!

nice