Determine the function so that the integral holds

[gop]

Homework Statement

Determine all functions H:(0,\infty)\to\mathbb{R}
so that H(x)=\frac{1}{x}\int_{2}^{x}t(3-2H'(t))\ dt,\ \ x>0

Homework Equations

The Attempt at a Solution

I tried to integrate by parts to change the integral into to the form of a differential equation, or at least an equation that could be differentiated to yield a differential equation. However; I always end up with a 1/x term in from of my integral thus making it impossible to get only H'(x) and H(x). Basically, I can't eliminate the integral of H(x) from my equation

 

[sutupidmath]

try to differentiate both parts with respect to x, and see if you can get anything. or just try to integrate the right-handed side and see if you can come up with sth.

 

[gop]

I had to multiply the equation with x then the differentiation works out fine
thx

 

[sutupidmath]

Yeah, after that you should have gotten a differential equation, i think it should be a linear one, which is solved using an integrating factor. So all solutions of that diff. eq are indeed all the functions H(t)=(3/4) x+Cx^-3, if i can remember it well, because i did it yesterday or sth after you posted, and i am not sure if this is the exact answer i got.
But as a check to your answer take the derivative of this function, and plug it in the integral, integrate it and see if you get the same function H(t).

 

[gop]

yeah its a linear one and the result I got is H=(3/4)x + Cx^(-1/3)

 

[sutupidmath]

yeah its a linear one and the result I got is H=(3/4)x + Cx^(-1/3)

are u sure u didn't do any mistakes on your way!??i think the answer should read

H(x)=\frac{3}{4}x+Cx^{-3} and not

H(x)=\frac{3}{4}x+Cx^{\frac{-1}{3}

as a means of checking your answer, follow the instructions i gave u in post # 4

These kind of problems are quite interesting, this one too! ...lol...

 

[gop]

I tried in maple
dsolve(H(x)+x*(diff(H(x), x)) = x*(3-2*(diff(H(x), x))));

and got
H(x) = \frac{3}{4}x + \frac{_C1}{x^{1/3}}

so I guess the 1/3 is correct.
I did the exercise in class today the one thing I forgot was that because I differentiate I relax the constraints and have to require
H(2) = 0