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 . (t^2-5t+4)dx/dt=1,(t>4,x(9)=0)

Answer 1

dx = dt/(t² - 5t + 4); x = ∫1/3[1/(t - 4) - 1/(t - 1)]dt; x(t) = 1/3lnI(t - 4)/(t - 1)] + C; x(9) = 1/3ln(5/8) + C = 0;

C = - 1/3ln(5/8); x(t) = 1/3[ln(t - 4) - ln(t - 1)] - 1/3ln(5/8)