Chapter 7.3, Problem 16E

To determine

To find an initial value problem.

Answer to Problem 16E

  $\text{The required answer is, }y=\frac{4}{3}x{\left(x+2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{8}{15}{\left(x+2\right)}^{5/2}+\frac{28}{15}$

Explanation of Solution

Given information: $\text{The equation is},\frac{dy}{dx}=2x\sqrt{x+2}\text{ and }y=0\text{ when }x=-1.$

Calculation:

  $\begin{array}{l}\text{Find the antiderivative of}2x\sqrt{x+2}\text{ by using parts}\text{. It is usually better idea to}\\ \text{differentiate 2}x\text{ than to antidifferentiate it}\text{.}\end{array}$

  $\begin{array}{l}\frac{dy}{dx}=2x\sqrt{x+2}\\ dy=2x\sqrt{x+2}.dx\\ {\displaystyle \int dy={\displaystyle \int 2x\sqrt{x+2}.dx\left(\text{taking integration both side}\right)}}\end{array}$

  $\begin{array}{l}\text{ let }u=2x\text{ and }dv={\left(x+2\right)}^{1/2}dx,\\ du=2dx\text{ and }v=\frac{{\left(x+2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{3/2}=\frac{2}{3}{\left(x+2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$

  $y=2x.\frac{2}{3}{\left(x+2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\displaystyle \int \frac{2}{3}{\left(x+2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(2dx\right)}\left(\text{using integration by parts}\right)$

  $\begin{array}{l}y=\frac{4}{3}x{\left(x+2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{4}{3}{\displaystyle \int {\left(x+2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx}\\ y=\frac{4}{3}x{\left(x+2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{4}{3}\left(\frac{{\left(x+2\right)}^{5/2}}{5/2}\right)+C\\ y=\frac{4}{3}x{\left(x+2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{4}{3}\left(\frac{2}{5}{\left(x+2\right)}^{5/2}\right)+C\\ y=\frac{4}{3}x{\left(x+2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{8}{15}{\left(x+2\right)}^{5/2}+C\end{array}$

  $\text{using the initial condition, }y=0\text{ when }x=-1.$

  $\begin{array}{l}y=\frac{4}{3}x{\left(x+2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{8}{15}{\left(x+2\right)}^{5/2}+C\\ \text{ 0}=\frac{4}{3}\left(-1\right){\left(-1+2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{8}{15}{\left(-1+2\right)}^{5/2}+C\\ \text{ 0}=-\frac{4}{3}{\left(1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{8}{15}{\left(1\right)}^{5/2}+C\\ \text{ 0}=-\frac{4}{3}-\frac{8}{15}+C\\ \text{ C}=\frac{28}{15}\end{array}$

  $\begin{array}{l}\text{Thus, }\\ y=\frac{4}{3}x{\left(x+2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{8}{15}{\left(x+2\right)}^{5/2}+\frac{28}{15}\end{array}$

  

  $\begin{array}{l}\text{Figure shows a graph of this function superimposed in a slope field for }\\ \frac{dy}{dx}=2x\sqrt{x+2},\text{ to which it conforms nicely}\text{.}\\ \end{array}$