Suppose we have the following differential equation:
$$a)\;\; f(x) = 1 - x\int_\limits{1}^{x}f(t)dt$$
and a solution:
$$b) \;\; f(x) = xe^{(1 - x^2)/2} - xe^{-x^2/2}\int_\limits{1}^{x}t^{-2}e^{t^2/2}dt$$
which has the properties that (i) it is continuous for $x > 0$ and (ii) it satisfies (a).
I need to find all functions that satisfy both properties (continuous on $x > 0$, and solve (a)). However, my answer is wrong. Cannot figure why.
I define $y = \int_\limits{1}^x f(t)dt$. Then the part (a) becomes:
$$y' = 1 - xy \Rightarrow y' + xy = 1$$
Now, I assume that for the point $a = 1$, $y(1) = b$. I use the standard solution formula for the linear differential equation of the first order:
$$P(x) = x; Q(x) = 1; A(x) = \int_\limits{1}^xtdt = x^2/2 - 1/2$$
and the solution is:
$$y = be^{1/2 - x^2/2} + e^{1/2 - x^2/2}\int_\limits{1}^xe^{t^2/2 - 1/2}dt$$
If we differentiate this, we arrive at $y' = f(x)$, which is:
$$f(x) = y' = -bxe^{1/2 - x^2/2} - xe^{1/2 - x^2/2}\int_\limits{1}^xe^{t^2/2 - 1/2}dt + 1$$
But this answer does not give anything resembling (b), since the latter has $t^{-2}$ factor under the integral sign. Moreover, the book says that the function (b) is the ONLY solution, which my answer clearly does not agree with, since I have a constant $b$, which can be manipulated to give a different answers.
Can someone specify what is wrong in this proof (it is not finished yet, I seem stuck)? Is it possible to prove this by continuing this approach I am taking (you can show how)?