Question

Solve the differential equation.

dx/dt = 1 - t + x - tx

Answer 1

`\displaystyle(dx)/(dt) = 1 - t + x - tx\ \Rightarrow\ (dx)/(dt) = 1(1- t) + x(1 - t) \ \Rightarrow \\ \\ (dx)/(dt) = (1+x)(1 - t) \ \Rightarrow\ \int (dx)/(1+x) = \int (1 - t)\ dt\ \Rightarrow \\ \\ \textstyle \ln|1 + x| = t - (1)/(2)t^2 + C\ \Rightarrow\ |1 + x| = e^(t - t^2/2 + C )\ \Rightarrow \\ \\ 1 + x = \pm e^(t - t^2/2) \cdot e^C\Rightarrow \\ \\ x = -1 + Ke^(t - t^2/2),\ \text{where $K$ is any nonzero constant.}`