Question

Find a closed-form solution to the integral equation y(x) = 3 + Z x e dt ty(t) , x > 0. In other words, express y(x) as a function that doesn’t involve an integral. (Hint: Use the Fundamental Theorem of Calculus to obtain a differential equation. You can find an initial condition by evaluating the original integral equation at a strategic value of x.)

Answer 1

`y{x} = √(7+2Inx)`

Step-by-step explanation:

`y(x)= 3 + \int\limits^x_e {dx}/ \, ty(t) , x>0}`

Let say; By y(x)= y(e)

we have;

`y(e)= 3 + \int\limits^e_e {dt}/ \, ty= 3+0`

Using Fundamental Theorem of Calculus and differentiating by Lebiniz Rule:

`y^(1) (x) = 0 + 1/ xy`

`y^(1) = 1/xy`

dy/dx = 1/xy

`\int\limits {y} \, dxy = \int\limits \, dx/x`

`y^(2)/2 Inx + C`

RECALL: y(e) = 3

`(3)^(2) / 2 = In (e) + C`

`(9)/(2) =In(e)+C`

`(9)/(2) - 1 = C`

`(7)/(2) = C`

`y^(2) / 2 = In x +C`

`y^(2) / 2 = In x +7/2`

MULTIPLYING BOTH SIDE BY 2 , TO ELIMINATE THE DENOMINATOR, WE HAVE;

`y^(2) = {7+2Inx}`

`y{x} = √(7+2Inx)`