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

Answer:

$y{x} = \sqrt{7+2Inx}$

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  

$(3)^{2} / 2 = In (e) + C$  

$\frac{9}{2} =In(e)+C$  

$\frac{9}{2} - 1 = C$

$\frac{7}{2} = C$  

$y^{2} / 2 = In x +C$

$y^{2} / 2 = In x +7/2$

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

$y^{2} = {7+2Inx}$  

$y{x} = \sqrt{7+2Inx}$