Chapter 7.3, Problem 16E

To determine

To find : the solution of the differential equation.

Answer to Problem 16E

The solution of differential equation is $p={\left(\sqrt{t}\text{+}\sqrt{2}-1\right)}^2$ .

Explanation of Solution

Given information :

The differential equation is $\frac{dp}{dt}=\sqrt{pt},p\left(1\right)=2$ .

Calculation : here the equation is separable, so write the equation in the terms of differentials and then integrate both sides,

  $\begin{array}{l}\frac{dp}{dt}=\sqrt{pt}\\ \frac{dp}{dt}=\sqrt{p}·\sqrt{t}\text{ [simplify}]\\ \frac{dp}{\sqrt{p}}\text{=}\frac{dt}{\sqrt{t}}\text{ [separate the variables]}\\ {\displaystyle \int \frac{dp}{\sqrt{p}}=}{\displaystyle \int \frac{dt}{\sqrt{t}}\text{ [integrate both side]}}\\ 2\sqrt{p}\text{=}2\sqrt{t}\text{+}c\text{ [simplify integrate] }\\ \sqrt{p}\text{=}\sqrt{t}\text{+}\frac{c}{2}\text{ [divide by 2]}\\ p={\left(\sqrt{t}\text{+}\frac{c}{2}\right)}^2\text{ [square] }\end{array}$

Here $c$ is a arbitrary constant.

Initial condition is $p\left(1\right)=2$ , now substitute $t=1\text{ and }p=2$ ,

  $\begin{array}{l}p={\left(\sqrt{t}\text{+}\frac{c}{2}\right)}^2\\ 2={\left(\sqrt{1}\text{+}\frac{c}{2}\right)}^2\text{ [substitute]}\\ 2={\left(\text{1+}\frac{c}{2}\right)}^2\text{ [ sqaure]}\\ \sqrt{2}\text{=1+}\frac{c}{2}\text{ [sqaure root]}\\ \sqrt{2}-1=\frac{c}{2}\text{ [subtract 1 from both side]}\end{array}$

So the solution of equation is $p={\left(\sqrt{t}\text{+}\sqrt{2}-1\right)}^2$

Hence, the solution of differential equation is $p={\left(\sqrt{t}\text{+}\sqrt{2}-1\right)}^2$ .