Chapter 7.3, Problem 18E

To determine

To find : the solution of the differential equation.

Answer to Problem 18E

The solution of differential equation is $L=\frac{1}{t\left(1-\ln t\right)-2}$ .

Explanation of Solution

Given information :

The differential equation is $\frac{dL}{dt}=k{L}^2\ln t,L\left(1\right)=-1$ .

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

  $\begin{array}{l}\frac{dL}{dt}=k{L}^2\ln t\\ \frac{dL}{L^2}\text{=}\ln tdt\text{ [separate the variables]}\\ {\displaystyle \int \frac{dL}{L^2}=}{\displaystyle \int \ln tdt\text{ [integrate both side]}}\\ -\frac{1}{L}\text{=}t\left(\ln t-1\right)+c\text{ [simplify integrate] }\\ \frac{1}{L}\text{=}t\left(1-\ln t\right)+c\text{ [multiply by }-1\text{]}\\ L=\frac{1}{t\left(1-\ln t\right)+c}\text{ [cross multiplication]}\end{array}$

Here $c$ is a arbitrary constant.

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

  $\begin{array}{l}L=\frac{1}{t\left(1-\ln t\right)+c}\\ -1=\frac{1}{1\left(1-\ln 1\right)+c}\text{ [substitute]}\\ -1=\frac{1}{1\left(1-0\right)+c}\text{ [ln1=0}]\\ -1=\frac{1}{1+c}\text{ [subtrcat and multiply]}\\ \text{1}+c=-1\text{ [cross multiplication]}\\ c=-2\text{ [subtract }-1\text{ from both side]}\end{array}$

So the solution of equation is $L=\frac{1}{t\left(1-\ln t\right)-2}$

Hence, the solution of differential equation is $L=\frac{1}{t\left(1-\ln t\right)-2}$ .