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I was thinking about the problem that was as follows:

The integral equation $x(t)-\displaystyle \int_{0}^{1}[\cos (t) \sec (s) x(s)]ds=\sinh (t), 0\leq t\leq 1,$ has

(a)no solution,

(b)a unique solution,

(c)more than one but finitely many solution,

(d)infinitely many solutions.

My attempts: From the given equation,we get $$x(t)=\cos(t)\int_{0}^{1}\sec(s)x(s)ds + \sinh(t)=C \cos(t)+\sinh(t)$$ where $C=\int_{0}^{1}\sec(s)x(s)ds$ and so $$C=\int_{0}^{1}\sec(s)[\cos(s)C +\sinh(s)]ds=C\int_{0}^{1}ds+\int_{0}^{1}\sec(s)\sinh(s)ds$$ and hence we get, $\int_{0}^{1}\sec(s)\sinh(s)ds=0.$

From here,i could not progress.Am i going in the right direction? Please help.Thanks in advance for your time.

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1 Answer 1

$$x(t)=\cos t\int_0^1x(s)\sec s\,ds + \sinh t=C\cos t+\sinh t$$ where $C=\int_0^1x(s)\sec s\,ds$ and so $$C\cos t=\int_0^1[C\cos s+\sinh s]\sec s\,ds=C\int_0^1ds+\int_0^1\sec s\sinh s\,ds=C+k,$$ where $k=\int_0^1\sec s\sinh s\,ds\approx0.739$, so since no choice of $C$ will make this equation true for all $t$, there is no solution.

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