Q.(a) Solve : $(D^2-2D+1)y=e^{2x}+e^x$.

(a) $y=(A+Bx)e^{x}+e^{2x}+\tfrac{x^{2}}{2}e^{x}$. (b) $|Z|=1.77<1.96\Rightarrow$ accept the claim.

(a) Solve $(D^{2}-2D+1)y=e^{2x}+e^{x}$.

Complementary function: auxiliary equation $m^{2}-2m+1=(m-1)^{2}=0\Rightarrow m=1,1$, so $\text{CF}=(A+Bx)e^{x}$.

Particular integral for $e^{2x}$ (put $D=2$; $(D-1)^2=(2-1)^2=1 e0$):

$\text{PI}_1=\frac{1}{(D-1)^{2}}e^{2x}=\frac{1}{(2-1)^{2}}e^{2x}=e^{2x}.$

Particular integral for $e^{x}$ (here $D=1$ makes $(D-1)^2=0$, a double root, so multiply by $x^{2}$):

$\text{PI}_2=\frac{1}{(D-1)^{2}}e^{x}=\frac{x^{2}}{2!}e^{x}=\frac{x^{2}}{2}e^{x}.$

General solution:

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

(b) Test the ambulance claim (large-sample $Z$-test).

Claim mean $\mu=8.9$ min; sample $n=50,\ \bar X=9.3,\ s=1.6$.

• $H_0:\mu=8.9$ vs $H_1:\mu e8.9$ (two-tailed), $\alpha=0.05$. …