I've been pondering over a question sparked by two math problems: How can one assert from the outset that a problem contains hidden initial values, and what methods can be employed to discern this? My approach felt rather rudimentary; I found myself substituting various values into each new equation I derived, only to realize that the answers included a constant 'C' (lacking initial value information), which led to a significant waste of time on problem T63. This misconception made me believe that it was impossible to determine initial values for such problems, which in turn led me to overlook the implicit initial value information in problem T82, preventing me from defining 'C' and resulting in confusion.
Before any manipulation, the problems were stated as follows:
T63: $\int^1_{0}f(tx)dt = f(x)+xsinx\quad f(x) =\underline{\hspace{2cm}}$
T82: $\int_0^{x}f(t)dt = x+sinx+\int_0^{x}tf(x-t)dt \quad f(x) =\underline{\hspace{2cm}}$
What I am capable of doing includes separating the terms involving the limits of integration via substitution and solving differential equations, both of which were not problematic. Consequently, the problems could be simplified to:
After the first manipulation:
T63: $\int_0^{x}f(t)dt = xf(x)+x^2sinx$
T82: $\int_0^{x}f(t)dt = x+sinx+x\int_0^{x}f(t)dt-\int_0^{x}tf(t)dt$
Regardless of whether it was before or after the first manipulation, I tried numerous values but failed to discern any initial value information.
Upon a second manipulation, involving differentiation of both sides, I obtained:
T63: $f(x) = f(x)+xf'(x)+2xsinx+x^2 cosx$ $\rightarrow$ $f'(x) = -2sinx-xcosx \quad(x\neq 0)$
T82: $f(x) = 1+cosx+\int_0^{x}f(t)dt+xf(x)-xf(x)$ $\rightarrow$ $f(x) = 1+cosx+\int_0^{x}f(t)dt (x \in R)$
At this juncture, T63 presented a linear differential equation, which was straightforward to solve; interestingly, T82 revealed an initial value at the point zero, which could then be utilized to find a particular solution in subsequent steps.
However, the pressing question remains: Why do some equations of the form $F(\int_{0}^{x}f(t)dt,f(x),g(x))$=0 yield initial values while others do not, as illustrated by the first manipulation of T63 and the second manipulation of T82? Is there an underlying reason for this discrepancy? Surely, there must be a more efficient method than my current approach of incessantly substituting and testing values within each new equation derived. I am at a loss for explanations and would greatly appreciate any insights on this matter.
I have checked most of your manipulations and it all looks correct:
$$\tag*{T63} \textstyle\int_0^1f(tx)\,dt=f(x)+x\sin x $$ can easily be brought into the form $$\tag*{T63'} \textstyle\int_0^xf(t)\,dt=xf(x)+x^2\sin x\,. $$ With a bit more work (essentially differentiation and integration by parts) $$\tag*{T82} \textstyle\int_0^xf(t)\,dt =\textstyle x+\sin x+\int_0^xtf(x-t)\,dt $$ can be brought into the form $$ f(x) =\textstyle 1+\cos x+\int_0^xf(t)\,dt $$ which can be viewed as an inhomogeneous linear ODE with initial condition $f(0)=2\,.$ Its solution is $$ f(x) =\frac{3e^x+\sin x+\cos x}{2}\,. $$ Now to your pressing question: both problems are obviously of the form $F(\color{red}{x},\int_0^xf(t)\,dt,f(x),g(x))=0$ but for T63 we have $$ F(\color{red}{x},U,V,W)=-U+\color{red}{x}V+W\,, $$ while for T82 it is $$ F(\color{red}{x},U,V,W)=-U+V+W\,. $$ independent of $\color{red}{x}\,$. The difference is that the first $F$ does not specify $V=f(x)$ when we set $\color{red}{x}$ to zero.
To find solutions for T62 we differentiate its simple form T63' to get $$ f'(x)=-2\sin x+x\cos x\,. $$ (Seems like you or I have a sign error there.) This leads to $$ f(x)=C-\cos x+x\sin x\,,\quad\textstyle\int_0^xf(t)\,dt=Cx-x\cos x $$ which solves T63 for every constant $C\,.$