solve the initial-value problem for $x$ as a function of $t$.\n$(2t^3 - 2t^2 + t - 1) \\frac{dx}{dt} = 3…

solve the initial-value problem for $x$ as a function of $t$.\n$(2t^3 - 2t^2 + t - 1) \\frac{dx}{dt} = 3, x(2) = 0$

solve the initial-value problem for $x$ as a function of $t$.\n$(2t^3 - 2t^2 + t - 1) \\frac{dx}{dt} = 3, x(2) = 0$

Answer

Answer:

$x(t) = \ln|t-1| - \frac{1}{2}\ln(2t^2+1) + \frac{1}{2}\ln 9$