Chapter 7.3, Problem 15E

To determine

To find : the solution of the differential equation.

Answer to Problem 15E

The solution of differential equation is $\frac{y^2}{2}+\frac{1}{3}{\left(3+y^2\right)}^{\frac{3}{2}}\text{=}\ln x·\frac{x^2}{2}-\frac{x^3}{6}\text{+}\frac{10}{3}$ .

Explanation of Solution

Given information :

The differential equation is $x\ln x=y\left(1+\sqrt{3+y^2}\right)y\text{'}\text{, }y\left(1\right)=1$ .

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

  $\begin{array}{l}x\ln x=y\left(1+\sqrt{3+y^2}\right)y\text{'}\\ x\ln x=y\left(1+\sqrt{3+y^2}\right)\frac{dy}{dx}\text{ [}y\text{'}=\frac{dy}{dx}]\\ y\left(1+\sqrt{3+y^2}\right)dy\text{=}x\ln xdx\text{ [separate the variables]}\\ \left(y+y\sqrt{3+y^2}\right)dy=x\ln xdx\text{ [multiply]}\\ {\displaystyle \int \left(y+y\sqrt{3+y^2}\right)dy=}{\displaystyle \int x\ln xdx\text{ [integrate both side]}}\\ \frac{y^2}{2}+\frac{1}{2}·\frac{2}{3}{\left(3+y^2\right)}^{\frac{3}{2}}\text{=}\ln x·\frac{x^2}{2}-\frac{x^3}{6}\text{+}c\text{ [simplify integrate] }\\ \frac{y^2}{2}+\frac{1}{3}{\left(3+y^2\right)}^{\frac{3}{2}}\text{=}\ln x·\frac{x^2}{2}-\frac{x^3}{6}\text{+}c\text{ [simplify] }\end{array}$

Here $c$ is a arbitrary constant.

Initial condition is $y\left(1\right)=1$ , now substitute $x=1\text{ and }y=1$ ,

  $\begin{array}{l}\frac{y^2}{2}+\frac{1}{3}{\left(3+y^2\right)}^{\frac{3}{2}}\text{=}\ln x·\frac{x^2}{2}-\frac{x^3}{6}\text{+}c\\ \frac{{\left(1\right)}^2}{2}+\frac{1}{3}{\left(3+{\left(1\right)}^2\right)}^{\frac{3}{2}}\text{=}\ln \left(1\right)·\frac{{\left(1\right)}^2}{2}-\frac{{\left(1\right)}^3}{6}\text{+}c\text{ [substitute]}\\ \frac{1}{2}\text{+}\frac{1}{3}{\left(3+1\right)}^{\frac{3}{2}}\text{=0}-\frac{1}{6}+c\text{ [ln|1|=0 and sqaure]}\\ \frac{1}{2}\text{+}\frac{1}{3}{\left(4\right)}^{\frac{3}{2}}\text{=0}-\frac{1}{6}+c\text{ [add]}\\ \frac{1}{2}\text{+}\frac{8}{3}\text{=}-\frac{1}{6}+c\text{ [simplify]}\\ \frac{19}{6}=-\frac{1}{6}+c\text{ [add]}\\ \frac{19}{6}+\frac{1}{6}=-\frac{1}{6}+\frac{1}{6}+c\text{ [add }\frac{1}{6}\text{ both side]}\\ \frac{10}{3}=c\text{ [add and simplify]}\end{array}$

So the solution of equation is $\frac{y^2}{2}+\frac{1}{3}{\left(3+y^2\right)}^{\frac{3}{2}}\text{=}\ln x·\frac{x^2}{2}-\frac{x^3}{6}\text{+}\frac{10}{3}$

Hence, the solution of differential equation is $\frac{y^2}{2}+\frac{1}{3}{\left(3+y^2\right)}^{\frac{3}{2}}\text{=}\ln x·\frac{x^2}{2}-\frac{x^3}{6}\text{+}\frac{10}{3}$ .