$EI \, y'' = \frac{1}{2}Px - P\langle x - \frac{1}{2}L \rangle$
$EI \, y' = \frac{1}{4}Px^2 - \frac{1}{2}P\langle x - \frac{1}{2}L \rangle^2 + C_1$
$EI \, y = \frac{1}{12}Px^3 - \frac{1}{6}P\langle x - \frac{1}{2}L \rangle^3 + C_1x + C_2$
At x = 0, y = 0, therefore, C2 = 0
At x = L, y = 0
$0 = \frac{1}{12}PL^3 - \frac{1}{6}P\langle L - \frac{1}{2}L \rangle^3 + C_1L$
$0 = \frac{1}{12}PL^3 - \frac{1}{48}PL^3 + C_1L$
$C_1 = -\frac{1}{16}PL^2$
Thus,
$EI \, y = \frac{1}{12}Px^3 - \frac{1}{6}P\langle x - \frac{1}{2}L \rangle^3 - \frac{1}{16}PL^2x$
Maximum deflection will occur at x = ½ L (midspan)
$EI \, y_{max} = \frac{1}{12}P(\frac{1}{2}L)^3 - \frac{1}{6}P (\frac{1}{2}L - \frac{1}{2}L)^3 - \frac{1}{16}PL^2 (\frac{1}{2}L)$
$EI \, y_{max} = \frac{1}{96}PL^3 - 0 - \frac{1}{32}PL^3$
$y_{max} = -\dfrac{PL^3}{48EI}$
The negative sign indicates that the deflection is below the undeformed neutral axis.
Therefore,
$\delta_{max} = \dfrac{PL^3}{48EI}$ answer
