Integral equations

(FB/Math/BM – Suggested by Souradeep P./Sachin Modi respectively)

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Find f(x) such that  f(x) = x + ∫ [0,1] (x2y + xy2) f(y) dy.

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The x-factors can be moved outside the integral on the right hand side:

f(x) = x + x²·∫ [0,1] y·f(y) dy + x·∫ [0,1] y²·f(y) dy

The integrals on the right hand side are constants. Thus (A, B constants)

f(x) = Ax² + Bx

Insert this into (*) to find A and B:

(**) Ax² + Bx = x + ∫ [0,1] (x²y + xy²)(Ay² + By) dy

∫ [0,1] (x²y + xy²)(Ay² + By) dy =
= ∫ [0,1] Ax²y³ + Bx²y² + Axy⁴ + Bxy³ dy =
= [Ax²y⁴/4 + Bx²y³/3 + Axy⁵/5 + Bxy⁴/4]|[y=0,1] =
= Ax²/4 + Bx²/3 + Ax/5 + Bx/4

Thus, by (**)

Ax² + Bx = x + Ax²/4 + Bx²/3 + Ax/5 + Bx/4 for all x. Insert x = ±1:

A + B = 1 + A/4 + B/3 + A/5 + B/4
A – B = -1 + A/4 + B/3 – A/5 – B/4

which gives A = 80/119, B = 180/119 and so

f(x) = 20x(4 x + 9)/119

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f ‘(x) = f(x) + ∫ [0,1] f (y) dy

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(1) f'(x)=f(x)+∫[0,1]f(y) dy
f(x) = A·eˣ + B, where B = ∫[0,1]f(y) dy
Insert into (1):
A·eˣ = A·eˣ + B + ∫[0,1] A·e^y + B dy
A·eˣ = A·eˣ + B + [A·e^y + By]|[0,1]
…A·eˣ = A·eˣ + B + Ae + B – A
B + Ae + B – A = 0, so B = A(e – 1)/2. Thus

f(x) = A·(eˣ + (e – 1)/2) where A is an arbitrary constant.

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