From the figure below
$EI \, y'' = \frac{1}{2}w_o Lx - w_o x ( \frac{1}{2} x)$
$EI \, y'' = \frac{1}{2}w_o Lx - \frac{1}{2}w_o x^2$
$EI \, y' = \frac{1}{4}w_o Lx^2 - \frac{1}{6}w_o x^3 + C_1$
$EI \, y = \frac{1}{12}w_o Lx^3 - \frac{1}{24}w_o x^4 + C_1x + C_2$
At x = 0, y = 0, therefore C2 = 0
At x = L, y = 0
$0 = \frac{1}{12}w_o L^4 - \frac{1}{24}w_o L^4 + C_1L$
$C_1 = -\frac{1}{24}w_o L^3$
Therefore,
$EI \, y = \frac{1}{12}w_o Lx^3 - \frac{1}{24}w_o x^4 - \frac{1}{24}w_o L^3 x$
Maximum deflection will occur at x = ½ L (midspan)
$EI \, y_{max} = \frac{1}{12}w_o L (\frac{1}{2}L)^3 - \frac{1}{24}w_o (\frac{1}{2}L)^4 - \frac{1}{24}w_o L^3 (\frac{1}{2}L)$
$EI \, y_{max} = \frac{1}{96}w_o L^4 - \frac{1}{384}w_o L^4 - \frac{1}{48}w_o L^4$
$EI \, y_{max} = -\frac{5}{384}w_o L^4$
$\delta_{max} = \dfrac{5w_o L^4}{384EI}$ answer
Taking W = woL:
$\delta_{max} = \dfrac{5(w_o L)(L^3)}{384EI}$
$\delta_{max} = \dfrac{5W L^3}{384EI}$ answer
