Chapter 7, Problem 6RE

To determine

To solve : the differential equation

Answer to Problem 6RE

The solution of the differential equation is $x=-1+A{e}^{t-\frac{t^2}{2}}$

Explanation of Solution

Given information : The differential equation is $\frac{dx}{dt}=1-t+x-tx$

Calculation :

The differential equation is $\frac{dx}{dt}=1-t+x-tx$

Rewrite the differential equation $\frac{dx}{dt}=\left(1-t\right)\left(1+x\right)$

  $\frac{dx}{1+x}=\left(1-t\right)dt$ Assuming $x\ne -1$

  $\begin{array}{l}{\displaystyle \int \frac{dx}{1+x}={\displaystyle \int \left(1-t\right)}}\\ \ln \left|1+x\right|=t-\frac{t^2}{2}+C\\ 1+x=\pm e^C{e}^{t-\frac{t^2}{2}}=A{e}^{t-\frac{t^2}{2}}\\ x=-1+A{e}^{t-\frac{t^2}{2}}\end{array}$

Note that $x=-1$ is also a solution of the differential equation

Thus, the general solution is $x=-1+A{e}^{t-\frac{t^2}{2}}$

Where $A$ is arhibitory constant