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## Homework Statement

seperate and solve using partial fractions

dx/dt=9-4x

^{2}, x(0)=0

## The Attempt at a Solution

rearranging gives dx/(9-4x

^{2}) = 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...