Chapter 7.3, Problem 26E

To determine

To solve : the differential equation and graph several members of the family of solutions.

Answer to Problem 26E

The solution of differential equation is $e^y\left(y-1\right)=\frac{1}{3}{\left(x^2+1\right)}^{\frac{3}{2}}+c$ .

Explanation of Solution

Given information :

The differential equation is $y\text{'}=\frac{x\sqrt{x^2+1}}{y{e}^y}$ .

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

  $\begin{array}{l}y\text{'}=\frac{x\sqrt{x^2+1}}{y{e}^y}\\ \frac{dy}{dx}=\frac{x\sqrt{x^2+1}}{y{e}^y}\text{ [}y\text{'}=\frac{dy}{dx}]\\ y{e}^{y}dy=x\sqrt{x^2+1}dx\text{ [separate the variables]}\\ {\displaystyle \int y{e}^{y}dy=}{\displaystyle \int x\sqrt{x^2+1}dx\text{ [integrate both side]}}\\ e^y\left(y-1\right)=\frac{1}{2}·\frac{2}{3}{\left(x^2+1\right)}^{\frac{3}{2}}+c\text{ [simplify integrate] }\\ e^y\left(y-1\right)=\frac{1}{3}{\left(x^2+1\right)}^{\frac{3}{2}}+c\text{ [divide]}\end{array}$

Here $c$ is a arbitrary constant.

Hence, the solution of differential equation is $e^y\left(y-1\right)=\frac{1}{3}{\left(x^2+1\right)}^{\frac{3}{2}}+c$ .

Now substitute different value for c to get different members of family of solutions.

  $\begin{array}{l}{e}^y\left(y-1\right)=\frac{1}{3}{\left(x^2+1\right)}^{\frac{3}{2}}+1\text{ [}c=1]\\ e^y\left(y-1\right)=\frac{1}{3}{\left(x^2+1\right)}^{\frac{3}{2}}+2\text{ [}c=2]\\ e^y\left(y-1\right)=\frac{1}{3}{\left(x^2+1\right)}^{\frac{3}{2}}+3\text{ [}c=3]\\ e^y\left(y-1\right)=\frac{1}{3}{\left(x^2+1\right)}^{\frac{3}{2}}-2\text{ [}c=-2]\\ e^y\left(y-1\right)=\frac{1}{3}{\left(x^2+1\right)}^{\frac{3}{2}}-1\text{ [}c=-1]\end{array}$

Using implicit plotting capability of a CAS to graph the curves $e^y\left(y-1\right)=\frac{1}{3}{\left(x^2+1\right)}^{\frac{3}{2}}+1,e^y\left(y-1\right)=\frac{1}{3}{\left(x^2+1\right)}^{\frac{3}{2}}+2,e^y\left(y-1\right)=\frac{1}{3}{\left(x^2+1\right)}^{\frac{3}{2}}+3,e^y\left(y-1\right)=\frac{1}{3}{\left(x^2+1\right)}^{\frac{3}{2}}-2,e^y\left(y-1\right)=\frac{1}{3}{\left(x^2+1\right)}^{\frac{3}{2}}-1$ .

The graph is obtained as:

  

When constant c value positive and increases, the curve of the graph stretched upwards, and for negative values the curve of graph moving downwards.