Differential equation. separate variables and solve using partial fractions

[ProPatto16]

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 can't seem to manipulate my answer to get there...

 

[cragar]

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 .

 

[ProPatto16]

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 :(

 

[cragar]

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

 

[ProPatto16]

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!

 

[cragar]

nice