(a) Roots $5,5$: C.F. $=(A+Bx)e^{5x}$; P.I. $=2x^{2}e^{5x}+\tfrac15$. (b) $L=\frac{\sum p_1q_0}{\sum p_0q_0}\times100=116.60$, $P=\frac{\sum p_1q_1}{\sum p_0q_1}\times100=118.12$, $F=\sqrt{LP}=117.36$.

Part (a) — $(D^{2}-10D+25)y=4e^{5x}+5$

Step 1 — Complementary function. Auxiliary equation $m^{2}-10m+25=0\Rightarrow(m-5)^{2}=0\Rightarrow m=5,5$ (repeated):

$\text{C.F.}=(A+Bx)e^{5x}.$

Step 2 — Particular integral for $4e^{5x}$. Here $f(D)=(D-5)^{2}$ and $a=5$ is a double root, so use $\dfrac{1}{(D-5)^{2}}e^{5x}=\dfrac{x^{2}}{2!}e^{5x}$:

$\text{P.I.}_1=4\cdot\frac{x^{2}}{2}e^{5x}=2x^{2}e^{5x}.$

Step 3 — Particular integral for $5$ (i.e. $5e^{0x}$):

$\text{P.I.}_2=\frac{5}{D^{2}-10D+25}\bigg|_{D=0}=\frac{5}{25}=\frac{1}{5}.$

Step 4 — General solution:

$y=(A+Bx)e^{5x}+2x^{2}e^{5x}+\frac{1}{5}.$

Part (b) — Laspeyre's, Paasche's and Fisher's index numbers

Step 1 — Required sums:

Step 2 — Laspeyre's index (base-year weights): …