Chapter 6.1, Problem 20E

To determine

To solve: the given initial problem.

Answer to Problem 20E

  $s=t^3-t^2$ .

Explanation of Solution

Given:

  $\frac{ds}{dt}=t\left(3t-2\right)$ and $s=0$ when $t=1$ .

Concept used:

The general solution of the exact differential equation of the form $\frac{dy}{dx}=f\left(x\right)$ is $y={\displaystyle \int f\left(x\right)dx}$ .

Calculation:

Consider the given differential equation,

  $\frac{ds}{dt}=t\left(3t-2\right)$

Now the general solution is,

  $\begin{array}{c}s={\displaystyle \int t\left(3t-2\right)dt}\\ s={\displaystyle \int \left(3t^2-2t\right)dt}\\ s=3\cdot \frac{t^3}{3}-2\cdot \frac{t^2}{2}+C\\ s=t^3-t^2+C\end{array}$

Now by using the given initial condition, it follows that

  $\begin{array}{c}0={\left(1\right)}^3-{\left(1\right)}^2+C\\ 0=1-1+C\\ C=0\end{array}$

Hence the solution is,

  $s=t^3-t^2$