Q.(a) State all the five properties of groups. OR

(a) lists and explains the defining axioms of a group; (b) independently solves the given linear ODE (complementary function + particular integral by cases) and checks the result against the claimed solution term by term.

(a) The five properties of a group

An algebraic structure $(G,*)$, where $*$ is a binary operation on the non-empty set $G$, is a group if it satisfies:

1. Closure axiom: for all $a,b\in G$, $a*b\in G$ (the operation never leaves the set).
2. Associative axiom: for all $a,b,c\in G$, $(a*b)*c=a*(b*c)$.
3. Identity axiom: there exists $e\in G$ such that $a*e=e*a=a$ for every $a\in G$ ($e$ is the identity element).
4. Inverse axiom: for every $a\in G$, there exists $a^{-1}\in G$ such that $a*a^{-1}=a^{-1}*a=e$.
5. Commutative (Abelian) axiom: for all $a,b\in G$, $a*b=b*a$. Note: properties 1–4 alone already make $(G,*)$ a group; property 5 is the additional condition that makes it specifically an Abelian group — it is listed here as the fifth defining property since the question asks for all five, but it is not required of a group in general.

(b) Solve $(5D^2-8D-4)y=5e^{-2x/5}+2e^x+3$ and verify

1. Complementary function: auxiliary equation $5m^2-8m-4=0\Rightarrow m=\dfrac{8\pm\sqrt{64+80}}{10}=\dfrac{8\pm12}{10}$, giving $m=2$ or $m=-\dfrac25$.
2. So $CF=Ae^{2x}+Be^{-2x/5}$ — matches the claimed solution's first two terms.
3. PI for $5e^{-2x/5}$: let $f(D)=5D^2-8D-4$, $f(m)=5m^2-8m-4$. Since $m=-\dfrac25$ is a root of the auxiliary equation, $f(-\tfrac25)=0$, so the usual $\dfrac{e^{ax}}{f(a)}$ rule fails and we use $\dfrac{x\,e^{ax}}{f'(a)}$ instead, where $f'(m)=10m-8$.
4. $f'\!\big(-\tfrac25\big)=10\big(-\tfrac25\big)-8=-4-8=-12$.
5. $\mathrm{PI}_1=5\cdot\dfrac{x\,e^{-2x/5}}{-12}=-\dfrac{5}{12}x\,e^{-2x/5}$ — matches the claimed $-\dfrac{5}{12}xe^{-2x/5}$ exactly. …