By symmetry
$R_1 = R_2 = bw_o$
$EI \, y'' = bw_o x - \frac{1}{2} w_o \langle x - a \rangle^2$
$EI \, y' = \frac{1}{2} bw_o x^2 - \frac{1}{6} w_o \langle x - a \rangle^3 + C_1$
$EI \, y = \frac{1}{6} bw_o x^3 - \frac{1}{24} w_o \langle x - a \rangle^4 + C_1x + C_2$
At x = 0, y = 0, therefore C2 = 0
At x = a + b, y' = 0
$0 = \frac{1}{2} bw_o (a + b)^2 - \frac{1}{6} w_o b^3 + C_1$
$C_1 = \frac{1}{6} w_o b^3 - \frac{1}{2} bw_o (a + b)^2$
Therefore,
$EI \, y = \frac{1}{6} bw_o x^3 - \frac{1}{24} w_o \langle x - a \rangle^4 + \frac{1}{6} w_o b^3x - \frac{1}{2} bw_o (a + b)^2x$
Maximum deflection will occur at x = a + b (midspan)
$EI \, y_{max} = \frac{1}{6} bw_o(a + b)^3 - \frac{1}{24} w_ob^4 + \frac{1}{6} w_ob^3(a + b) - \frac{1}{2} bw_o(a + b)^3$
$EI \, y_{max} = -\frac{1}{3} bw_o(a + b)^3 - \frac{1}{24} w_ob^4 + \frac{1}{6} w_ob^3(a + b)$
$EI \, y_{max} = -\frac{1}{24} w_ob \, [ \, 8(a + b)^3 + b^3 - 4b^2(a + b) \, ]$
Therefore,
$\delta_{max} = \dfrac{w_ob}{24EI}[ \, 8(a + b)^3 + b^3 - 4b^2(a + b) \, ]$ answer
Checking:
When a = 0, 2b = L, thus b = ½ L
$\delta_{max} = \dfrac{w_o(\frac{1}{2}L)}{24EI}[ \, 8(0 + \frac{1}{2}L)^3 + (\frac{1}{2}L)^3 - 4(\frac{1}{2}L)^2(0 + \frac{1}{2}L) \, ]$
$\delta_{max} = \dfrac{w_oL}{48EI}[ \, L^3 + \frac{1}{8}L^3 - \frac{1}{2}L^3 \, ]$
$\delta_{max} = \dfrac{w_oL}{48EI}[ \, \frac{5}{8}L^3 \, ]$
$\delta_{max} = \dfrac{5w_oL^4}{384EI}$ (okay!)
