Chapter 7.3, Problem 34E

To determine

To solve : the integral equation.

Answer to Problem 34E

the solution of integral equation is $y=\ln x^2+2$ .

Explanation of Solution

Given information :

The integral equation is $y\left(x\right)=2+{\displaystyle {\int }_1^x\frac{dt}{ty\left(t\right)}}$ .

Calculation: to solve the integral equation first find the initial condition of the equation,

For initial equation substitute $x=1,t=1$ in the integral equation,

  $\begin{array}{l}y\left(x\right)=2+{\displaystyle {\int }_1^x\frac{dt}{ty\left(t\right)}}\\ y\left(1\right)=2+{\displaystyle {\int }_1^1\frac{dt}{1·y\left(1\right)}}\text{ [substitute]}\\ y\left(1\right)=2\text{[simplify]}\end{array}$

Now solve the integral equation, differentiate the equation then it will be converted into differential equation then solve the differential equation,

  $\begin{array}{l}y\left(x\right)=2+{\displaystyle {\int }_1^x\frac{dt}{ty\left(t\right)}}\\ y\text{'}\left(x\right)=0+\frac{1}{xy\left(x\right)}\text{ [differentiate with respect to }x\text{ and put }t=x]\\ y\text{'}\left(x\right)=\frac{1}{xy\left(x\right)}\end{array}$

So the differential equation for integral equation is $y\text{'}\left(x\right)=\frac{1}{xy\left(x\right)}$ .

Now solve the differential equation,

Let $y\left(x\right)=y$

  $\begin{array}{l}y\text{'}\left(x\right)=\frac{1}{xy\left(x\right)}\\ y\text{'}=\frac{1}{xy}\text{ [subtitute }y\left(x\right)=y]\\ \frac{dy}{dx}=\frac{1}{xy}\text{ [}y\text{'}=\frac{dy}{dx}]\\ ydy=\frac{dx}{x}\text{ [separate variables]}\\ {\displaystyle \int ydy={\displaystyle \int \frac{dx}{x}}}\text{ [integrate both side]}\\ \frac{y^2}{2}=\ln \left|x\right|+c\text{ [simplify integrate]}\\ y=2\ln \left|x\right|+c\text{ [multiply by 2]}\\ y=\ln x^2+c\text{ [simplify}]\end{array}$

Where c are arbitrary constant.

Since $y\left(1\right)=2$ , so

  $\begin{array}{l}y=\ln x^2+c\\ 2=\ln {\left(1\right)}^2+c\text{ [substitue]}\\ 2=0+c\text{ [ln1=0]}\\ c=2\end{array}$

Thus the solution of integral equation is $y=\ln x^2+2$