Solve and find the flaw in this integral equation

Question

The following integral equation often appears in the books and it has once been asked in the prestigious examination called IIT JEE (M) dated 10-04-2016. The question is: $\forall x \in R-\{0\}$, if $y(x)$ is differentiable function such that $$ x\int_{1}^{x}~ y(t)~dt =(x+1) \int_{1}^{x} t ~y(t)~dt.$$ Find $y(x)$. Some four interesting expression of $y(x)$ were given as alternatives in this MCQ type question.

Solving this you may find a serious flaw in this question.

I would like to thank Ninad Sutrave for expressing a doubt about this question.

Answer 1

The flaw in the question is that $y(x) = 0$ obviously satisfies the equation and is everywhere differentiable. So without doing any math I can provide an answer.

Or at least, that seems to be the flaw. But $y(x) = 0$ is also the unique solution to the problem. To see this, consider a function $Y(x)$ that satisfies $Y''(x) = y(x)$. Do some FTC and integration by parts and you get $$ x^2 Y'(x) -(x+1)Y(x) = Y'(1) - (1+x)Y(1). $$ Differential equation solves to $$ Y(x) = C\,xe^{-1/x} + Y(1) + (x-1)Y'(1), $$ which can only be satisfied if $C = 0$. Thus, $y(x) = Y''(x) = 0$.

Differentiating both sides twice gives an equation that solves to $y(x) = Ce^{-1/x}/x^3$, but plugging that back into the original equation also gives $C = 0$.

So while the fact that this is the solution may be a flaw in the question, it does seem to be well-posed, in the sense that the solution is unique.

Answer 2

A commentary:

$$ \forall x \in R-\{0\}~~~~~~~~(1)$$ $$x \int_{1}^{x} y(t) ~dt =(x+1)~\int_{1}^{x} t~y(t)~dt.~~~~~(2)$$ D.w.r.t. $x$ $$\int_{1}^{x} y(t) ~dt +xy=\int_{1}^{x} t~ y(t) ~dt + (x+1) xy.~~~~~~(3)$$ Eliminating second term on RHS by (1), we get $$\int_{1}^{x} y(t)~dt= x^2(x+1) y(x). ~~~~~~(4)$$ From here it follows $$ y(1)=0 ~~~~~~~~(5)$$ Differentiate (4) w.r.t $x$, we get $$\int \frac{dy}{y} = \int \frac{1-3x^2-2x}{x^2+x^3}~dx +C.~~~~~(6)$$ We get $$ y(x) = \frac{C}{x^3}~ e^{-1/x}.~~~~~~~(7)$$ Resolution-1: In order to satisfy (5), $C=0$ so the solution is $y(x)=0$. Usually (5) is either ignored or not obtained, hence the wrong solution (7) is admitted. Let us now put (7) in (2), we get $$ LHS= (1+x) e^{-1/x} -\frac{2x}{C}, ~RHS= (1+x) e^{-1/x} -\frac{x+1}{C}.$$ These two sides can be equal only for one isolated point $x=1$. This contradicts the very first condition, namely, (1).

Resolution-2: This is an ill posed Integral equation which holds only for a isolated point $x=1$. One can only differentiate only an identity or a relation ship which holds for a continuous domain of $x$ e.g., $[a,b].$

In short, a given Integral must actually hold in a continuous domain $[a,b]$, otherwise the very first step of differentiation cannot be even initiated.