(i)$a=\mathrm{constant},$ $b=b\left(r\right)$, ${{\rm{e}}}^{-b}b^{\prime} r+2{{\rm{e}}}^{-b}-2=0$ implies $b=\mathrm{ln}\left(\tfrac{1}{1+{k}_{1}{r}^{2}}\right),$ Q = r and $T=\left(\tfrac{2}{{r}^{2}}+2{k}_{1}\right),$ where ${k}_{1}\,\in {\mathfrak{R}}\setminus \{0\}.$

(ii)$a=a\left(r\right)$, $b=\mathrm{constant},$ $2{ra}^{\prime\prime} +{ra}{{\prime} }^{2}-2a^{\prime} =0$ ⇒ $a\,=\mathrm{ln}{r}^{4},$ e^b = 1, Q = r and T = 10r⁻².

(iii)$a=a\left(r\right)$, $b=b\left(r\right)$, a = b⁻¹, ${r}^{2}\left(b^{\prime\prime} -b{{\prime} }^{2}\right)+2\left(1,-,{{\rm{e}}}^{b}\right)\,=0$ implies $b=\mathrm{ln}{\left(1-\tfrac{{k}_{1}}{r}+\tfrac{{k}_{2}{r}^{2}}{3}\right)}^{-1}$, $a=\mathrm{ln}\left(1-\tfrac{{k}_{1}}{r}+\tfrac{{k}_{2}{r}^{2}}{3}\right),$ Q = r and $T=\left(\tfrac{2}{{r}^{2}}+2{k}_{2}\right),$ where ${k}_{1},{k}_{2}\in {\mathfrak{R}}\setminus \{0\}.$

(iv)$a=a\left(r\right)$, $b=b\left(r\right)$, a = b⁻¹, ${{\rm{e}}}^{a}\left(\tfrac{a^{\prime\prime} }{2}+\tfrac{a{{\prime} }^{2}}{2}-\tfrac{1}{{r}^{2}}\right)+\tfrac{1}{{r}^{2}}\,=0$ ⇒ $a=\mathrm{ln}\left(1-\tfrac{2M}{r}\right)$, $b=\mathrm{ln}{\left(1-\tfrac{2M}{r}\right)}^{-1},$ Q = r and $T=\tfrac{2}{{r}^{2}},$ where M represents the Arnowitt–Deser–Misner mass.

(v)$a=a\left(r\right)$, $b=b\left(r\right)$, a = b⁻¹, ${r}^{2}\left(a^{\prime\prime} +a{{\prime} }^{2}\right)\,-2\left(1-{{\rm{e}}}^{-a}\right)=0$ implies $a=\mathrm{ln}\left(1-\tfrac{{\rm{\Lambda }}{r}^{2}}{3}\right)$, $b\,=\mathrm{ln}{\left(1-\tfrac{{\rm{\Lambda }}{r}^{2}}{3}\right)}^{-1},$ Q = r and $T=\left(\tfrac{2}{{r}^{2}}-2{\rm{\Lambda }}\right),$ where Λ is the cosmological constant.

(vi)$a=a\left(r\right)$, $b=b\left(r\right)$, ${ra}^{\prime} +1=0$ which gives $a=\mathrm{ln}\left(\tfrac{{k}_{1}}{r}\right)$, $4{{\rm{e}}}^{b}-{rb}^{\prime} +1=0$ implies $b=\mathrm{ln}\left(\tfrac{r}{{k}_{2}-4r}\right),$ Q = r and T = 0, where ${k}_{1},{k}_{2}\in {\mathfrak{R}}\setminus \{0\}.$

(vii)$a=a\left(r\right)$, $b=b\left(r\right)$, ${ra}^{\prime} -2=0$ implies $a=\mathrm{ln}\left({k}_{1}{r}^{2}\right)$, ${{\rm{e}}}^{b}-{rb}^{\prime} -2=0$ ⇒ $b=\mathrm{ln}\left(\tfrac{2}{1+2{k}_{2}{r}^{2}}\right),$ Q = r and $T\,=\displaystyle \frac{3\left(1+2{k}_{2}{r}^{2}\right)}{{r}^{2}},$ where ${k}_{1},{k}_{2}\in {\mathfrak{R}}\setminus \{0\}.$

(viii)$a=a\left(r\right)$, $b=b\left(r\right)$, $\left(\tfrac{a^{\prime\prime} }{2}+\tfrac{a{{\prime} }^{2}}{4}\right)=0$ ⇒ $a\,=\mathrm{ln}{\left(\tfrac{{k}_{1}r+{k}_{2}}{2}\right)}^{2}$, $-{k}_{1}b^{\prime} {{\rm{e}}}^{-b}+2\left({k}_{1}r+{k}_{2}\right)=0$ which implies $b=\mathrm{ln}\left(\tfrac{{k}_{1}}{{k}_{1}{k}_{3}-{k}_{1}{r}^{2}-2{k}_{2}r}\right),$ Q = 1 and T = 0, where ${k}_{1},{k}_{2},{k}_{3}\in {\mathfrak{R}}\setminus \{0\}.$

(ix)$a=a\left(r\right)$, $b=b\left(r\right)$, a = b⁻¹, $a^{\prime\prime} +a{{\prime} }^{2}+2{{\rm{e}}}^{-a}=0$ implies $a=\mathrm{ln}\left({k}_{2}-{k}_{1}r-{r}^{2}\right)$, $b=\mathrm{ln}{\left({k}_{2}-{k}_{1}r-{r}^{2}\right)}^{-1},$ Q = 1 and T = 0, where ${k}_{1},{k}_{2}\in {\mathfrak{R}}\setminus \{0\}.$

(x)$a=\mathrm{constant}\,=\,{k}_{1}\ne 0$, $b=\mathrm{constant}\,=\,{k}_{2}\ne 0$, $a=\mathrm{ln}\left({k}_{1}\right)$, $b=\mathrm{ln}\left({k}_{2}\right)$, ${QQ}^{\prime\prime} -Q{{\prime} }^{2}+{k}_{2}=0$ implies $Q=r\sqrt{{k}_{2}}$ and $T=\tfrac{2}{{k}_{2}{r}^{2}},$ where ${k}_{1},{k}_{2}\in {\mathfrak{R}}\setminus \{0\}$ with k₁ ≠ k₂.

Now, we discuss the second possibility arising from equation (9), that is $T^{\prime} =0$, giving

(xi)$a=a\left(r\right)$, $b=b\left(r\right)$, ${ra}^{\prime\prime} -a^{\prime} =0$ implies $a\,=\left(\tfrac{{k}_{1}{r}^{2}}{2}+{k}_{2}\right)$, ${rb}^{\prime} \left({ra}^{\prime} +1\right)+2=0$ ⇒ $b=\mathrm{ln}{\left(\tfrac{{k}_{3}\sqrt{{k}_{1}{r}^{2}+1}}{r}\right)}^{2},$ Q = r and $T=\tfrac{2}{{k}_{3}^{2}},$ where ${k}_{1},{k}_{2},{k}_{3}\in {\mathfrak{R}}\left({k}_{1},{k}_{3}\ne 0\right).$

(xii) $a=a\left(r\right)$, $b=b\left(r\right)$, $1+{rb}^{\prime} =0$ implies $b=\mathrm{ln}\left(\tfrac{{k}_{1}}{r}\right)$, ${r}^{2}a^{\prime\prime} -{rb}^{\prime} -2=0$ ⇒ $a=\mathrm{ln}\left(\tfrac{{k}_{3}{{\rm{e}}}^{{k}_{2}r}}{r}\right),$ Q = r and $T=\tfrac{2{k}_{2}}{{k}_{3}},$ where ${k}_{1},{k}_{2},{k}_{3}\in {\mathfrak{R}}\left({k}_{1},{k}_{3}\ne 0\right).$

(xiii)$a=a\left(r\right)$, $b=b\left(r\right)$, $2+{rb}^{\prime} =0$ implies $b=\mathrm{ln}\left(\tfrac{{k}_{1}}{{r}^{2}}\right)$, ${ra}^{\prime\prime} -a^{\prime} \left(1+{rb}^{\prime} \right)=0$ ⇒ $a=\mathrm{ln}\left({k}_{3}{r}^{{k}_{2}}\right),$ Q = r and $T=\tfrac{2\left({k}_{2}+1\right)}{{k}_{1}},$ where ${k}_{1},{k}_{2},{k}_{3}\in {\mathfrak{R}}\left({k}_{1},{k}_{2},{k}_{3}\ne 0\right).$

(xiv)$a=a\left(r\right)$, $b=b\left(r\right)$, $2+{ra}^{\prime} =0$ implies $a=\mathrm{ln}\left(\tfrac{{k}_{1}}{{r}^{2}}\right)$, ${ra}^{\prime\prime} -b^{\prime} \left(1+{ra}^{\prime} \right)=0$ ⇒ $b=\mathrm{ln}\left(\tfrac{{k}_{2}}{{r}^{2}}\right),$ Q = r and $T=\tfrac{-2}{{k}_{2}},$ where ${k}_{1},{k}_{2}\in {\mathfrak{R}}\left({k}_{1},{k}_{2}\ne 0\right).$

(xv)$a=a\left(r\right)$, $b=b\left(r\right)$, r²a″ − 2 = 0 implies $a=\mathrm{ln}\left(\tfrac{{{\rm{e}}}^{{k}_{1}r+{k}_{2}}}{{r}^{2}}\right)$, $a^{\prime} +b^{\prime} \left(1+{ra}^{\prime} \right)=0$ ⇒ $b=\mathrm{ln}\left[\tfrac{{k}_{3}\left({k}_{1}r-1\right)}{{r}^{2}}\right],$ Q = r and $T=\tfrac{2}{{k}_{3}},$ where ${k}_{1},{k}_{2},{k}_{3}\in {\mathfrak{R}}\left({k}_{1},{k}_{3}\ne 0\right).$

(xvi)$a=\mathrm{constant}$, $b=b\left(r\right)$, ${rb}^{\prime} +2=0$ ⇒ $b=\mathrm{ln}\left(\tfrac{{k}_{1}}{{r}^{2}}\right)$, e^a =1, Q = r and $T=\tfrac{2}{{k}_{1}},$ where ${k}_{1}\in {\mathfrak{R}}\setminus \{0\}.$

(xvii)$a=a\left(r\right)$, $b\,=\,\mathrm{constant}\,=\,{k}_{1}\ne 0$, ${r}^{2}a^{\prime\prime} -{ra}^{\prime} -2=0$ ⇒$a=\left(\tfrac{{k}_{2}{r}^{2}}{2}-\mathrm{ln}r+{k}_{3}\right),$ Q = r and $T=\tfrac{2{k}_{2}}{{{\rm{e}}}^{{k}_{1}}},$ where ${k}_{1},{k}_{2},{k}_{3}\in {\mathfrak{R}}\left({k}_{1},{k}_{2}\ne 0\right).$

(xviii)$a=a\left(r\right)$, $b=b\left(r\right)$, a″ = 0 implies $a=\left({k}_{1}r+{k}_{2}\right)$, ${ra}^{\prime} \left(1+{rb}^{\prime} \right)+{rb}^{\prime} +2\,=\,0$ ⇒ $b=\mathrm{ln}\left[\tfrac{{k}_{3}\left({k}_{1}r+1\right)}{{r}^{2}}\right],$ Q = r and $T=\tfrac{2}{{k}_{3}},$ where ${k}_{1},{k}_{2},{k}_{3}\in {\mathfrak{R}}\left({k}_{1},{k}_{3}\ne 0\right).$

(xix)$a=\mathrm{constant}\,=\,{k}_{1}\ne 0$, $b=b\left(r\right)$, Q″ = 0 implies $Q\,=\left({k}_{2}r+{k}_{3}\right)$, $b^{\prime} Q+2Q^{\prime} =0$ ⇒ $b=\mathrm{ln}\left[\tfrac{{k}_{4}}{{\left({k}_{2}r+{k}_{3}\right)}^{2}}\right]$ and $T\,=\tfrac{2{k}_{2}^{2}}{{k}_{4}},$ where ${k}_{1},{k}_{2},{k}_{3},{k}_{4}\in {\mathfrak{R}}\left({k}_{1},{k}_{2},{k}_{4}\ne 0\right).$

(xx)$a=\mathrm{constant}\,=\,{k}_{1}\ne 0$, $b=\mathrm{constant}\,=\,{k}_{2}\ne 0$, $a\,=\mathrm{ln}({k}_{1})$, $b=\mathrm{ln}({k}_{2})$, ${QQ}^{\prime\prime} -Q{{\prime} }^{2}=0$ implies $Q={{\rm{e}}}^{{k}_{3}r+{k}_{4}}$ and $T=\tfrac{2{k}_{3}^{2}}{{{\rm{e}}}^{{k}_{2}}},$ where ${k}_{1},{k}_{2},{k}_{3},{k}_{4}\in {\mathfrak{R}}\left({k}_{3}\ne 0\right)$ with k₁ ≠ k₂.

(a)The CVFs in cases (iv), (v), (vi), (viii), (ix), (xi), (xii), (xiii), (xiv), (xv), (xvii), (xviii) and (xix) become the KVFs. Physically, the KVFs produce conservation laws i.e. conservation of energy and linear momentum are related with the translational KVFs ∂_t and ∂_φ respectively, whereas conservation of angular momentum has been depicted by the rotational isometries Y₂ and Y₃ respectively.

(b)In cases (vii), (xvi) and (xx), the space-times admit proper CVFs. The dimension of CVFs for these cases has turned out to be six, of which four are KVFs which are given in equation (3), whereas the remaining two are proper CVFs which are ${Y}_{5}^{* * }$ and ${Y}_{6}^{* * }.$ The CVFs have a wide scope of application in loop quantum cosmology and astrophysics giving models of compact stars, dense stars and gravastars [85].

(c)In case (x), the space-time admits proper HVFs due to the conformal factor ξ being a non-zero constant. The HVFs are quite important from several points of view. First, HVFs have been found useful in discussing the constant of motion that allows for examining particle trajectories in space-time [84]. Secondly, the homothetic motion helps to address the singularity issue in GR. It is to be noted that by studying self-similar solutions of the EFEs, the class of HVFs has also played a significant role.