Chapter 7.1, Problem 24E

To determine

To solve the initial value problem using fundamental theorem.

Answer to Problem 24E

The solution of differential equation is $y={\displaystyle \underset{0}{\overset{s}{\int }}\sqrt[3]{\tan t}}dt+\text{ 4}$ .

Explanation of Solution

Given information:

It is given that

  $G^{\prime }\left(s\right)=\sqrt[3]{\tan s}\text{ and G}\left(0\right)=4$

Calculation:

According to fundamental theorem ${\displaystyle \underset{a}{\overset{s}{\int }}{G}^{\prime }\left(s\right)}dt=\text{ G}\left(s\right)-G\left(a\right)$ .This can also be written as $G\left(s\right)={\displaystyle \underset{a}{\overset{s}{\int }}{G}^{\prime }\left(t\right)}dt+\text{ G}\left(a\right)$ . The initial condition given is $G\left(0\right)=4$ .

Let $a=0$ , then $G\left(a\right)=G\left(0\right)=4$ , also it is given that $G^{\prime }\left(s\right)=\sqrt[3]{\tan s}$ .

Substituting $G\left(s\right)=y$ , $G\left(a\right)=G\left(0\right)=4$ and $G^{\prime }\left(s\right)=\sqrt[3]{\tan s}$

therefore

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{s}{\int }}{G}^{\prime }\left(s\right)}dt=\text{ G}\left(s\right)-G\left(a\right)\\ y={\displaystyle \underset{0}{\overset{s}{\int }}\sqrt[3]{\tan t}}dt+\text{ 4}\end{array}$