For the deflection of a clamped-free beam with a sinusoidal pile, we wish to show that the solution \[ y(x) = \frac{f}{24EI}x^2(x^2-4Lx+6L^2) - \frac{pgL}{EI\pi} \left( \frac{L^3}{\pi^3} \sin\frac{\pi}{L}x - \frac{x^3}{6} + \frac{L}{2}x^2 - \frac{L^2}{\pi^2}x\right) \] satsifies the Euler-Bernoulli beam equation (where \( f=-480wdg \) is the weight of the beam), \[ EIy''''(x) = f(x) = f - pg \sin\frac{\pi}{L}x \] and the clamped-free boundary conditions, \[ y(0) = y'(0) = y''(L) = y'''(L) = 0 \]

Let us begin by multiplying both sides of the solution by \( EI \), then deriving to get to \( EIy''''(x) \):

\[ EIy(x) = \frac{f}{24}(x^4-4Lx^3+6L^2x^2) - pg\frac{L}{\pi}\left(\frac{L^3}{\pi^3}\sin\frac{\pi}{L}x- \frac{x^3}{6}+\frac{L}{2}x^2-\frac{L^2}{\pi^2}x\right) \]
\[ EIy'(x) = \frac{f}{24}(4x^3 - 12Lx^2 + 12L^2x) - pg\frac{L}{\pi} \left( \frac{L^2}{\pi^2}\cos\frac{\pi}{L}x-\frac{x^2}{2}+Lx-\frac{L^2}{\pi^2} \right) \]
\[ EIy''(x) = \frac{f}{24}(12x^2 - 24Lx + 12L^2) - pg\frac{L}{\pi} \left( -\frac{L}{\pi}\sin\frac{\pi}{L}x-x+L \right) \]
\[ EIy'''(x) = \frac{f}{24}(24x - 24L) - pg\frac{L}{\pi} \left( -\cos\frac{\pi}{L}x - 1 \right) \]
\[ EIy''''(x) = f - pg\sin\frac{\pi}{L}x \]

We can see from the final equation that the Euler-Bernoulli beam equation is satisfied.
Evaluating the first four equations at the appropriate values proves the boundary conditions:

\[ EIy(0) = \frac{f}{24}(0^4-4L0^3+6L^20^2) - pg\frac{L}{\pi} \left( \frac{L^3}{\pi^3}\sin0- \frac{0^3}{6}+\frac{L}{2}0^2-\frac{L^3}{\pi^3}0 \right) = 0 \therefore y(0) = 0 \]
\[ EIy'(0) = \frac{f}{24}(4\cdot0^3-12L0^2+12L^20) - pg\frac{L}{\pi} \left(\frac{L^2}{\pi^2}\cos0- \frac{0^2}{2}+L0-\frac{L^2}{\pi^2} \right) = -pg\frac{L}{\pi} \left( \frac{L^2}{\pi^2}- \frac{L^2}{\pi^2} \right) = 0 \therefore y'(0) = 0 \]
\[ EIy''(L) = \frac{f}{24}(12L^2-24L^2+12L^2) - pg\frac{L}{\pi} \left( -\frac{L}{\pi}\sin\pi-L+L \right) = \frac{f}{24}(0) - pg\frac{L}{\pi}(0) = 0 \therefore y''(L) = 0 \]
\[ EIy'''(L) = \frac{f}{24}(24L-24L) - pg\frac{L}{\pi}(-\cos\pi - 1) = \frac{f}{24}(0)-pg\frac{L}{\pi}(0) = 0 \therefore y'''(L) = 0 \]

Thus the solution is correct.

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