Chapter 7.3, Problem 33E

To determine

To solve : the integral equation.

Answer to Problem 33E

the solution of integral equation is $y=1+e^{\frac{x^2}{2}}$ .

Explanation of Solution

Given information :

The integral equation is $y\left(x\right)=2+{\displaystyle {\int }_2^x[t-ty\left(t\right)}]dt$ .

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

For initial equation substitute $x=0,t=0$ in the integral equation,

  $\begin{array}{l}y\left(x\right)=2+{\displaystyle {\int }_2^x[t-ty\left(t\right)}]dt\\ y\left(0\right)=2+{\displaystyle {\int }_2^x[0-0·y\left(0\right)]}dt\text{ [substitute]}\\ y\left(0\right)=2\text{[simplify]}\end{array}$

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

  $\begin{array}{l}y\left(x\right)=2+{\displaystyle {\int }_2^x[t-ty\left(t\right)}dt\\ y\text{'}\left(x\right)=0+x-xy\left(x\right)\text{ [differentiate with respect to }x\text{ and put }t=x]\\ y\text{'}\left(x\right)=x-xy\left(x\right)\end{array}$

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

Now solve the differential equation,

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

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

Where A, c are arbitrary constant.

Since $y\left(0\right)=2$ , so

  $\begin{array}{l}y=1-A{e}^{\frac{x^2}{2}}\\ 2=1-A{e}^0\text{ [}y\left(0\right)=2]\\ 2=1-A\text{ [}{e}^0=1]\\ A=-1\text{ [simplify]}\end{array}$

Thus the solution of integral equation is $y=1+e^{\frac{x^2}{2}}$