Chapter 6.4, Problem 10E

To determine

To solve: the given initial value problem by using the separation of variables.

Answer to Problem 10E

  $y={\left(\ln x\right)}^4$

The solution is valid in the interval $\left(0,\infty \right)$ .

Explanation of Solution

Given:

  $\frac{dy}{dx}=\frac{4\sqrt{y}\ln x}{x}$ and $y=1$ when $x=e$ .

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,

  $\frac{dy}{dx}=\frac{4\sqrt{y}\ln x}{x}$

Now by separating the variables, it follows that

  $\frac{dy}{\sqrt{y}}=\frac{4\ln x}{x}dx$

Now by integrating on both sides, it follows that

  $\begin{array}{c}{\displaystyle \int \frac{dy}{\sqrt{y}}}={\displaystyle \int \frac{4\ln x}{x}dx}\\ 2\sqrt{y}=4{\displaystyle \int \frac{\ln x}{x}dx}\to \left(1\right)\end{array}$

Put $u=\ln x$ . Then,

  $\begin{array}{c}\frac{du}{dx}=\frac{1}{x}\\ du=\frac{1}{x}dx\end{array}$

Therefore, equation (1) becomes

  $\begin{array}{c}2\sqrt{y}=4{\displaystyle \int udu}\\ \sqrt{y}=2\cdot \frac{u^2}{2}+C\\ \sqrt{y}={\left(\ln x\right)}^2+C\end{array}$

Now by using the given initial conditions, it follows that

  $\begin{array}{c}\sqrt{1}={\left(\ln e\right)}^2+C\\ 1=1+C\\ C=0\end{array}$

Now put $C=0$ in $\sqrt{y}={\left(\ln x\right)}^2+C$ , it follows that

  $\begin{array}{c}\sqrt{y}={\left(\ln x\right)}^2+0\\ y={\left(\ln x\right)}^4\end{array}$

Clearly, the solution is valid in the interval $\left(0,\infty \right)$ .