Q.(a) Solve : $(3D^2+D-14)y=4-13e^{-\frac{7}{3}x}$

(a) Solve $(3D^2+D-14)y=4-13e^{-7x/3}$: $y=Ae^{2x}+Be^{-7x/3}-\tfrac27+xe^{-7x/3}$. (b) Binomial $n=4,p=0.18$: $0.397,\ 0.452,\ 0.980$.

Part (a) — second-order linear ODE with constant coefficients.

Complementary function. Auxiliary equation $3m^{2}+m-14=0$:

$m=\frac{-1\pm\sqrt{1+168}}{6}=\frac{-1\pm13}{6}\ \Rightarrow\ m=2\ \text{or}\ m=-\frac{7}{3}.$

$\text{C.F.}=A e^{2x}+B e^{-7x/3}.$

Particular integral for $4$ (i.e. $4e^{0\cdot x}$, put $D=0$): $f(0)=3(0)+0-14=-14$,

$\text{P.I.}_1=\frac{4}{3D^{2}+D-14}\bigg|_{D=0}=\frac{4}{-14}=-\frac{2}{7}.$

Particular integral for $-13e^{-7x/3}$ (put $D=-\tfrac73$): $f\!\left(-\tfrac73\right)=3\cdot\tfrac{49}{9}-\tfrac73-14=\tfrac{49}{3}-\tfrac73-14=14-14=0$ (resonance). Use $\dfrac{x}{f'(D)}$ with $f'(D)=6D+1$, $f'\!\left(-\tfrac73\right)=6\left(-\tfrac73\right)+1=-14+1=-13$:

$\text{P.I.}_2=-13\cdot\frac{x\,e^{-7x/3}}{f'(-7/3)}=-13\cdot\frac{x\,e^{-7x/3}}{-13}=x\,e^{-7x/3}.$

General solution:

$y=\text{C.F.}+\text{P.I.}_1+\text{P.I.}_2=A e^{2x}+B e^{-7x/3}-\frac{2}{7}+x\,e^{-7x/3}.$

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