Solve for $f(x)$ the integral equation $int_1^xf(t)dt=x(f(x))^2$ ?

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Answer 1 · 2016-11-19T23:20:12

$f(x)=1-1/sqrtx$ and $f(x)=0$

$int_1^xf(t)dt=x(f(x))^2$ deriving both sides

$f(x)=(f(x))^2+2xf(x)f'(x)$ or

$(2xf'(x)+f(x)-1)f(x)=0$ Supposing that $f(x) ne 0$ we have

$2xf'(x)+f(x)-1=0$ Solving this linear differential equation we obtain

$f(x)=1+C_1/sqrt(x)$ The constant is determined submiting this solution into the first relationship giving

$(x + 2(sqrt[x]-1)C_1-1)/x=(1+C_1/sqrtx)^2$ giving $C_1=-1$ so the function is

$f(x)=1-1/sqrtx$ and also $f(x)=0$

Solve for f(x)f(x) the integral equation int_1^xf(t)dt=x(f(x))^2int_1^xf(t)dt=x(f(x))^2 ?