Differential equation. seperate variables and solve using partial fractions!

  • Thread starter ProPatto16
  • Start date
  • #1
326
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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...
 

Answers and Replies

  • #2
2,544
3
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 .
 
  • #3
326
0
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 :(
 
  • #4
2,544
3
multiply both side by (3+2x) then get all the x's on one side and then factor out the x
 
  • #5
326
0
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!
 
  • #6
2,544
3
nice
 

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