Chapter 6, Problem 26RE

To determine

To solve: the given initial value problem.

Answer to Problem 26RE

  $y=\frac{x^3}{3}+2x-\frac{1}{x}-\frac{1}{3}$ .

Explanation of Solution

Given:

  $\frac{dy}{dx}={\left(x+\frac{1}{x}\right)}^2,y\left(1\right)=1$ .

Concept used:

A differential equation of the form $\frac{dy}{dx}=f\left(y\right)g\left(x\right)$ is called separable.

The variables can be separated by writing it in the form $\frac{1}{f\left(y\right)}dy=g\left(x\right)dx$ .

The solution is found by anti differentiating each side with respect to its thusly isolated variable.

Calculation:

Consider the given differential equation,

  $\frac{dy}{dx}={\left(x+\frac{1}{x}\right)}^2$

By simplification, it follows that

  $\begin{array}{l}\frac{dy}{dx}=x^2+2\cdot x\cdot \frac{1}{x}+\frac{1}{x^2}\\ \frac{dy}{dx}=x^2+2+x^{-2}\end{array}$

Now by separating the variables, it follows that

  $dy=\left(x^2+2+x^{-2}\right)dx$

Now by integrating on both sides, it follows that

  $\begin{array}{c}{\displaystyle \int dy}={\displaystyle \int \left(x^2+2+x^{-2}\right)dx}\\ y=\frac{x^3}{3}+2x+\frac{x^{-2+1}}{-2+1}+C\\ y=\frac{x^3}{3}+2x-\frac{1}{x}+C\end{array}$

Now by using the given initial condition, it follows that

  $\begin{array}{c}1=\frac{1^3}{3}+2\left(1\right)-\frac{1}{1}+C\\ 1=\frac{1}{3}+2-1+C\\ C=-\frac{1}{3}\end{array}$

Now put $C=-\frac{1}{3}$ in $y=\frac{x^3}{3}+2x-\frac{1}{x}+C$ , it follows that

  $y=\frac{x^3}{3}+2x-\frac{1}{x}-\frac{1}{3}$