Discussion
The solution of the integral equation
\[ \frac{\pi t}{2}=\int_{0}^{t}\frac{1}{\sqrt{t-s}}\,x(s)\,ds \]
is :
Answer: B
The solution is \(x(t) = \sqrt{t}\).
This is Abel's integral equation. Test \(x(s) = \sqrt{s}\) with the substitution \(s = tu\):
\[ \begin{aligned} \int_{0}^{t} \frac{\sqrt{s}}{\sqrt{t - s}}\,ds &= t\int_{0}^{1} u^{1/2}(1 - u)^{-1/2}\,du \\ &= t\,B\left(\tfrac{3}{2}, \tfrac{1}{2}\right) = t \cdot \frac{\pi}{2} \end{aligned} \]
since
\[ \begin{aligned} B\left(\tfrac{3}{2}, \tfrac{1}{2}\right) &= \frac{\Gamma(3/2)\Gamma(1/2)}{\Gamma(2)} \\ &= \frac{\sqrt{\pi}}{2} \cdot \sqrt{\pi} = \frac{\pi}{2} \end{aligned} \]
That matches the left-hand side \(\frac{\pi t}{2}\).
By the same computation, \(x = 1\) gives \(2\sqrt{t}\), \(x = t\) gives \(\frac{4}{3}t^{3/2}\), and \(x = \frac{1}{\sqrt{t}}\) gives the constant \(\pi\); none equals \(\frac{\pi t}{2}\).
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