Chapter 7.1, Problem 22E

To determine

To solve the initial value problem using fundamental theorem.

Answer to Problem 22E

The solution of differential equation is $u={\displaystyle \underset{0}{\overset{x}{\int }}\sqrt{2+\cos t}}dt-3$ .

Explanation of Solution

Given information:

It is given that

  $\frac{du}{dx}=\sqrt{2+\cos x}\text{ and }u=-3\text{ when }x=0$

Calculation:

According to fundamental theorem ${\displaystyle \underset{u}{\overset{x}{\int }}{f}^{\prime }\left(x\right)}dt=f\left(x\right)-f\left(a\right)$ .This can also be written as $f\left(x\right)={\displaystyle \underset{u}{\overset{x}{\int }}{f}^{\prime }\left(t\right)}dt+f\left(a\right)$ . The initial condition given is $u=-3$ when $x=0$ .

Let $a=0$ , then $f\left(a\right)=f\left(0\right)=-3$ , also it is given that $f^{\prime }\left(x\right)=\frac{du}{dx}=\sqrt{2+\cos x}$ .

Substituting $f\left(x\right)=u$ , $f\left(a\right)=f\left(0\right)=-3$ and $f^{\prime }\left(x\right)=\sqrt{2+\cos x}$

therefore

  $\begin{array}{l}f\left(x\right)={\displaystyle \underset{a}{\overset{x}{\int }}{f}^{\prime }\left(t\right)}dt+f\left(a\right)\\ u={\displaystyle \underset{0}{\overset{x}{\int }}\sqrt{2+\cos t}}dt-3\end{array}$