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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.)

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Answer:

[tex]y{x} = \sqrt{7+2Inx}[/tex]

Step-by-step explanation:

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

Let say; By y(x)= y(e)  

we have;  

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

Using Fundamental Theorem of Calculus and differentiating by Lebiniz Rule:

[tex]y^{1} (x) = 0 + 1/ xy[/tex]

[tex]y^{1} = 1/xy[/tex]  

dy/dx = 1/xy  

[tex]\int\limits {y} \, dxy = \int\limits \, dx/x[/tex]

[tex]y^{2}/2 Inx + C[/tex]

RECALL: y(e) = 3  

[tex](3)^{2} / 2 = In (e) + C[/tex]  

[tex]\frac{9}{2} =In(e)+C[/tex]  

[tex]\frac{9}{2} - 1 = C[/tex]

[tex]\frac{7}{2} = C[/tex]  

[tex]y^{2} / 2 = In x +C[/tex]

[tex]y^{2} / 2 = In x +7/2[/tex]

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

[tex]y^{2} = {7+2Inx}[/tex]  

[tex]y{x} = \sqrt{7+2Inx}[/tex]

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