Solve the initial-value problem for x as a function of t .

Question

Solve the initial-value problem for  x as a function of t .

(2t^3−2t^2+t−1)dx/dt=3, x(2)=0

Answer 1

Separate the variables:

dx=[3/(2t³-2t²+t-1)] dt

This integrates by partial fractions x+C=∫dx/(x-1)-2∫[(x-1)/(2x²+1)] dx

Answer 2

(2t^3−2t^2+t−1)dx/dt=3, x(2)=0

(dx/3)=dt/(2t^3−2t^2+t−1)=dt/{(t-1)(2t²+1)]

(dx/3)=[[(1/4)/(t-1)]-[(1/2)(t+1)/(2t²+1)]]dt

integrating

(1/3)x=(1/4)ln(t-1)-(1/8)ln(2t²+1)-(√2/4)tan^-1((√2)t)+C

t=2, x=0

C=(1/8)ln(9)+(√2/4)tan^-1(2√2)