$EI \, y'' = -\dfrac{M}{a}x + \dfrac{M}{a} \, \langle x - a \rangle$
$EI \, y' = -\dfrac{M}{2a}x^2 + \dfrac{M}{2a} \, \langle x - a \rangle^2 + C_1$
$EI \, y = -\dfrac{M}{6a}x^3 + \dfrac{M}{6a} \, \langle x - a \rangle^3 + C_1x + C_2$
At x = 0, y = 0, therefore C2 = 0
At x = a, y = 0
0 = -(M / 6a)(a3) + aC1
C1 = Ma / 6
Therefore,
$EI \, y' = -\dfrac{M}{2a}x^2 + \dfrac{M}{2a} \, \langle x - a \rangle^2 + \dfrac{Ma}{6}$
$EI \, y = -\dfrac{M}{6a}x^3 + \dfrac{M}{6a} \, \langle x - a \rangle^3 + \dfrac{Ma}{6}x$
Slope at x = a + b
$EI \, y' = -\dfrac{M}{2a}(a + b)^2 + \dfrac{M}{2a}(b^2) + \dfrac{Ma}{6}$
$EI \, y' = -\dfrac{M}{2a}(a^2 + 2ab + b^2) + \dfrac{M}{2a}(b^2) + \dfrac{Ma}{6}$
$EI \, y' = -\frac{1}{2}Ma - Mb - \dfrac{Mb^2}{2a} + \dfrac{Mb^2}{2a} + \frac{1}{6}Ma$
$EI \, y' = -\frac{1}{3}Ma - Mb$
$EI \, y' = -\frac{1}{3}M \, (a + 3b)$
$EI \, y' = -\frac{1}{3}M \, [ \, (a + b) + 2b \, ]$
$EI \, y' = -\frac{1}{3}M \, (L + 2b)$ answer
Deflection at x = a + b
$EI \, y = -\dfrac{M}{6a}(a + b)^3 + \dfrac{M}{6a}(b^3) + \dfrac{Ma}{6}(a + b)$
$EI \, y = -\dfrac{M}{6a}(a^3 + 3a^2b + 3ab^2 + b^3) + \dfrac{Mb^3}{6a} + \dfrac{Ma}{6}(a + b)$
$EI \, y = -\dfrac{Ma^2}{6} - \dfrac{Mab}{2} - \dfrac{Mb^2}{2} - \dfrac{Mb^3}{6a} + \dfrac{Mb^3}{6a} + \dfrac{Ma^2}{6} + \dfrac{Mab}{6}$
$EI \, y = -\frac{1}{3}Mab - \frac{1}{2}Mb^2$
$EI \, y = -\frac{1}{6}Mb \,(2a + 3b)$
$EI \, y = -\frac{1}{6}Mb \,[ \, 2(a + b) + b \, ]$
$EI \, y = -\frac{1}{6}Mb \,(2L + b)$
$EI \, \delta = \frac{1}{6}Mb \,(2L + b)$ answer
