### 3.3Problem 7.2 part (e) in Textbook

Starting with\begin{align} q^{\prime }\left ( t\right ) +\frac{\left ( I_{1}-I_{3}\right ) }{I_{2}}\Omega \ r\left ( t\right ) & =0\tag{1}\\ r^{\prime }\left ( t\right ) +\frac{\left ( I_{2}-I_{1}\right ) }{I_{3}}\Omega \ q\left ( t\right ) & =0\tag{2} \end{align}

To decouple the ODE’s, take derivatives and substituting back, we ﬁnd\begin{align} q^{\prime \prime }\left ( t\right ) +k\ q\left ( t\right ) & =0\tag{3}\\ r^{\prime \prime }\left ( t\right ) +k\ r\left ( t\right ) & =0\tag{4} \end{align}

Where $$k=\frac{\left ( I_{1}-I_{3}\right ) \left ( I_{2}-I_{2}\right ) }{I_{2}I_{3}}\Omega ^{2}$$. The solution to the above is (for stability $$k>0$$ )\begin{align} q\left ( t\right ) & =A\cos \sqrt{k}t+B\sin \sqrt{k}t\tag{5}\\ r\left ( t\right ) & =C\cos \sqrt{k}t+D\sin \sqrt{k}t\tag{6} \end{align}

Now, $$q\left ( 0\right ) =\frac{\text{Imp}}{I_{yy}}$$, hence $$A=\frac{\text{Imp}}{I_{yy}}$$ . To ﬁnd $$B$$ take derivative$$q^{\prime }\left ( t\right ) =-\sqrt{k}\frac{\text{Imp}}{I_{yy}}\sin \sqrt{k}t+B\sqrt{k}\cos \sqrt{k}t\tag{7}$$ But at $$t=0$$ then we go back and use Eq. (1) to ﬁnd $$q^{\prime }\left ( 0\right )$$, and equate the result to the above at $$t=0$$. Eq (1), at $$t=0$$, gives$q^{\prime }\left ( 0\right ) +\frac{\left ( I_{1}-I_{3}\right ) }{I_{2}}\Omega \ r\left ( 0\right ) =0$ But $$r\left ( 0\right ) =0\,$$, hence $$q^{\prime }\left ( 0\right ) =0$$, and so this results in $$B=0$$ in Eq.(7), hence the solution for $$q\left ( t\right )$$ is$q\left ( t\right ) =\frac{\text{Imp}}{I_{yy}}\cos \sqrt{k}t$ We do the same for $$r\left ( t\right ) .$$ From Eq. (6) we ﬁnd that $$C=0$$ since $$r\left ( 0\right ) =0$$, so now $$r\left ( t\right )$$ reduces to$r\left ( t\right ) =D\sin \sqrt{k}t$ Hence$$r^{\prime }\left ( t\right ) =\sqrt{k}D\cos \sqrt{k}t\tag{8}$$ To ﬁnd $$D$$, then we go back and use Eq. (2) to ﬁnd $$r^{\prime }\left ( 0\right )$$, and equate the result to the above at $$t=0$$. Eq (2), at $$t=0$$, gives$r^{\prime }\left ( 0\right ) +\frac{\left ( I_{2}-I_{1}\right ) }{I_{3}}\Omega \ q\left ( 0\right ) =0$ But $$q\left ( 0\right ) =\frac{\text{Imp}}{I_{yy}}$$, hence$r^{\prime }\left ( 0\right ) =-\frac{\left ( I_{2}-I_{1}\right ) }{I_{3}}\Omega \ \frac{\text{Imp}}{I_{yy}}$ Therefore, equate the above to Eq. (8) evaluated at $$t=0\,\$$gives$D=-\frac{\left ( I_{2}-I_{1}\right ) }{I_{3}\sqrt{k}}\Omega \ \frac{\text{Imp}}{I_{yy}}$ Hence Eq. (6) becomes$r\left ( t\right ) =\left ( -\frac{\left ( I_{2}-I_{1}\right ) }{I_{3}\sqrt{k}}\Omega \ \frac{\text{Imp}}{I_{yy}}\right ) \sin \sqrt{k}t$ so $$r\left ( t\right )$$ is sinusoidal as well.