## 2 Theory
\begin{array}{c}a_{1}=1,a_{2}=\frac{3(k-1)}{(k+1)}a_{1},a_{3}=\frac{3k}{(k+ 2)}a_{2}+\frac{\big{\{}\overline{D}_{4}-3(k-2)(k+1)\big{\}}}{(k+2)(k+1)}a_{1}, \\ a_{4}=\frac{3(k+1)}{(k+3)}a_{3}+\frac{\big{\{}\overline{D}_{4}-3(k-1)(k+2)\big{\} }}{(k+3)(k+2)}a_{2}+\frac{(k-1)\big{\{}(k-2)(k+3)+\overline{D}_{3}\big{\}}+ \overline{D}_{5}}{(k+3)(k+2)(k+1)}a_{1},\\ a_{5}=\frac{3(k+2)}{(k+4)}a_{4}+\frac{\big{\{}\overline{D}_{4}-3(k+3)\big{\}}}{( k+4)(k+3)}a_{3}+\frac{\big{\{}\overline{k}(k-1)(k+4)+\overline{D}_{3}\big{\}}+ \overline{D}_{5}}{(k+4)(k+3)(k+2)}a_{2}+\frac{k\big{\{}\overline{D}_{2}(k-1) -2\overline{D}_{5}\big{\}}-\overline{D}_{6}}{(k+4)(k+3)(k+2)(k+1)}a_{1}.\end{array}  \tag{18}\]  
The bending moment and shear force, obtained by substituting Equations (11) and (15) into Equations (7) and (9) can be rewritten as follows:  
\[ \begin{array}{c}M(\zeta)=N_{1}\Big{\{}(1-\zeta)^{3}W^{\prime\prime}-\overline {D}_{7}W^{\prime}\Big{\}},\end{array}  \tag{19}\]  
\[ \begin{array}{c}V(\zeta)=-N_{2}\Big{[}(1-\zeta)^{3}W^{\prime\prime\prime}-3( 1-\zeta)^{2}W^{\prime\prime}-\Big{\{}\overline{D}_{8}-\overline{D}_{1}(1- \zeta)^{3}\Big{\}}W^{\prime}-\big{(}\overline{D}_{9}-\overline{D}_{10}\big{)} W\Big{]},\end{array}  \tag{20}\]  
\[ \begin{array}{c}\text{where }N_{1}=\frac{E_{10}c^{2}}{L^{2}},N_{2}=\frac{E_{10}c^{3}}{L^{ 3}},\overline{D}_{7}=\frac{K_{60}L}{E_{10}c},\overline{D}_{8}=D_{3},\overline {D}_{9}=\frac{K_{W1}L^{3}}{E_{10}c^{3}},\overline{D}_{10}=\frac{m_{p1}\omega^{ 2}L^{3}}{E_{10}c^{3}}.\end{array} \]  
For \(n=2\), the nondimensional differential equation of a double-tapered beam is given by  
\[ \begin{array}{c}\big{(}\zeta^{4}-4\zeta^{3}+6\zeta^{2}-4\zeta+1\big{)}W^{ \prime\prime\prime\prime}+8\big{(}\zeta^{3}-3\zeta^{2}+3\zeta-1\big{)}W^{ \prime\prime\prime}\\ +\big{(}\overline{D}_{1}\zeta^{4}+\overline{D}_{2}\zeta^{3}+\overline{D}_{3} \zeta^{2}+\overline{D}_{4}\zeta+\overline{D}_{5}\big{)}W^{\prime\prime}\\ +\big{(}\overline{D}_{6}\zeta^{3}-3\overline{D}_{6}\zeta^{2}+3\overline{D}_{6} \zeta-\overline{D}_{6}\big{)}W^{\prime}+\big{(}\overline{D}_{7}\zeta^{2}-2 \overline{D}_{7}\zeta+\overline{D}_{7}\big{)}W=0,\end{array}  \tag{21}\]  
where \(\overline{D}_{1}=D_{2}\), \(\overline{D}_{2}=-4D_{2}\), \(\overline{D}_{3}=6(D_{2}+2)\), \(\overline{D}_{4}=-4(D_{2}+6)\), \(\overline{D}_{5}=(12+D_{2}-D_{3})\),By substituting Equation (15) into Equation (21), the indicial equation for \(n=2\) becomes Equation (16) for \(n=1\), and the recurrence relationship can be obtained as:  
\[ \begin{array}{l}a_{i+6}(k)\\ =\frac{4(k+i+3)}{(k+i+5)}a_{i+5}(k)-\frac{6(k+i+4)(k+i+1)+\overline{D}_{5}}{(k +i+4)(k+i+5)}a_{i+4}(k)\\ -\frac{\left(\overline{D}_{4}-4(k+i+5)(k+i)\right)(k+i+1)-\overline{D}_{6}}{(k +i+3)(k+i+4)(k+i+5)}a_{i+3}(k)\\ -\frac{(k+i+1)\left[(k+i+6)(k+i-1)+\overline{D}_{3}\right\}(k+i+3)+3\overline{ D}_{6}]+\overline{D}_{7}}{(k+i+2)(k+i+3)(k+i+4)(k+i+5)}a_{i+2}(k)\\ -\frac{(k+i)\left\{\overline{D}_{2}(k+i-1)-3\overline{D}_{6}\right\}-2\overline {D}_{7}}{(k+i+2)(k+i+3)(k+i+4)(k+i+5)}a_{i+1}(k)\\ -\frac{(k+i-1)\left\{\overline{D}_{6}+\overline{D}_{1}(k+i-2)\right\}+ \overline{D}_{6}}{(k+i+2)(k+i+3)(k+i+4)(k+i+5)}a_{i}(k).\end{array}  \tag{22}\]  
The Frobenius coefficients are as follows:  
\[