91Ó°ÊÓ

The given differential equation is,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{θ}\right)\mathbf{+}\mathbf{16}\mathbf{y}\left(\mathbf{θ}\right)\mathbf{=}\mathbf{tan}\left(\mathbf{4}\mathbf{θ}\right)                              ......\left(\mathbf{1}\right)$

Write the homogeneous differential equation of the equation (1),

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{θ}\right)\mathbf{+}\mathbf{16}\mathbf{y}\left(\mathbf{θ}\right)\mathbf{=}\mathbf{0}$

The auxiliary equation for the above equation ${\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{16}\mathbf{=}\mathbf{0}.$

Solve the auxiliary equation,

$\begin{array}{rcl}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{16}& \mathbf{=}& \mathbf{0}\\ {\mathbf{m}}^{\mathbf{2}}& \mathbf{=}& \mathbf{-}\mathbf{16}\\ \mathbf{m}& \mathbf{=}& \mathbf{Â\pm }\sqrt{\mathbf{-}\mathbf{16}}\\ \mathbf{m}& \mathbf{=}& \mathbf{Â\pm }\mathbf{4}\mathbf{i}\\ & & \end{array}$

The roots of the auxiliary equation are ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{i}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{i}$.

The complementary solution of the given equation is${\mathbf{y}}_{\mathbf{c}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{cos}\left(\mathbf{4}\mathbf{θ}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sin}\left(\mathbf{4}\mathbf{θ}\right)$.

Assume, the particular solution of equation (1),

${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{Acos}\left(\mathbf{4}\mathbf{θ}\right)\mathbf{+}\mathbf{Bsin}\left(\mathbf{4}\mathbf{θ}\right)                          ......\left(\mathbf{2}\right)$

Find the Wronskian $\mathbf{W}\left(\mathbf{cos}\mathbf{4}\mathbf{θ}\mathbf{,} \mathbf{sin}\mathbf{4}\mathbf{θ}\right)$

$\begin{array}{rcl}W\left(cos\left(4\mathrm{θ}\right),sin\left(4\mathrm{θ}\right)\right)& =& \left|\begin{array}{cc}cos\left(4\mathrm{θ}\right)& sin\left(\mathrm{θ}\right)\\ -4sin\left(4\mathrm{θ}\right)& 4cos\left(4\mathrm{θ}\right)\end{array}\right|\\ & =& 4{\cos }^{2}4\mathrm{θ}+4{\sin }^{2}4\mathrm{θ}\\ & =& 4\end{array}$

Now use the variation of parameters to find the value of A and B,

$\begin{array}{rcl}\mathbf{A}& \mathbf{=}& \mathbf{-}∫\frac{{\mathbf{y}}_{\mathbf{2}}\mathbf{g}\left(\mathbf{θ}\right)}{\mathbf{W}\left({\mathbf{y}}_{\mathbf{1}}\mathbf{,} {\mathbf{y}}_{\mathbf{2}}\right)}\mathbf{»åθ}\mathbf{;}  \mathbf{B}\mathbf{=}∫\frac{{\mathbf{y}}_{\mathbf{1}}\mathbf{g}\left(\mathbf{θ}\right)}{\mathbf{W}\left({\mathbf{y}}_{\mathbf{1}}\mathbf{,} {\mathbf{y}}_{\mathbf{2}}\right)}\mathbf{»åθ}\\ \mathbf{A}& \mathbf{=}& \mathbf{-}∫\frac{\mathbf{sin}\mathbf{4}\mathbf{θ³Ù²¹²Ô}\mathbf{4}\mathbf{θ}}{\mathbf{W}\left(\mathbf{cos}\mathbf{4}\mathbf{θ}\mathbf{,} \mathbf{sin}\mathbf{4}\mathbf{θ}\right)}\mathbf{»åθ}\mathbf{;}  \mathbf{B}\mathbf{=}∫\frac{\mathbf{cos}\mathbf{4}\mathbf{θ³Ù²¹²Ô}\mathbf{4}\mathbf{θ}}{\mathbf{W}\left(\mathbf{cos}\mathbf{4}\mathbf{θ}\mathbf{,} \mathbf{sin}\mathbf{4}\mathbf{θ}\right)}\mathbf{»åθ}\\ \mathbf{A}& \mathbf{=}& \mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}∫\mathbf{sin}\mathbf{4}\mathbf{θ³Ù²¹²Ô}\mathbf{4}\mathbf{θ»åθ}\mathbf{;}  \mathbf{B}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}∫\mathbf{cos}\mathbf{4}\mathbf{θ³Ù²¹²Ô}\mathbf{4}\mathbf{θ»åθ}\\ \mathbf{A}& \mathbf{=}& \mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}∫\frac{{\mathbf{sin}}^{\mathbf{2}}\mathbf{4}\mathbf{θ}}{\mathbf{cos}\mathbf{4}\mathbf{θ}}\mathbf{»åθ}\mathbf{;}  \mathbf{B}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}∫\mathbf{sin}\mathbf{4}\mathbf{θ»åθ}\\ \mathbf{A}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{16}}\mathbf{sin}\mathbf{4}\mathbf{θ}\mathbf{-}\frac{\mathbf{1}}{\mathbf{16}}\mathbf{ln}\left(\mathbf{sec}\mathbf{4}\mathbf{θ}\mathbf{+}\mathbf{tan}\mathbf{4}\mathbf{θ}\right)\mathbf{;}  \mathbf{B}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{16}}\mathbf{cos}\mathbf{4}\mathbf{θ}\\ & & \end{array}$

Therefore, the particular solution of equation (1),

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{Acos}\left(\mathbf{4}\mathbf{θ}\right)\mathbf{+}\mathbf{Bsin}\left(\mathbf{4}\mathbf{θ}\right) \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\left[\frac{\mathbf{1}}{\mathbf{16}}\mathbf{sin}\mathbf{4}\mathbf{θ}\mathbf{-}\frac{\mathbf{1}}{\mathbf{16}}\mathbf{ln}\left(\mathbf{sec}\mathbf{4}\mathbf{θ}\mathbf{+}\mathbf{tan}\mathbf{4}\mathbf{θ}\right)\right]\mathbf{cos}\mathbf{4}\mathbf{θ}\mathbf{+}\left[\mathbf{-}\frac{\mathbf{1}}{\mathbf{16}}\mathbf{cos}\mathbf{4}\mathbf{θ}\right]\mathbf{sin}\mathbf{4}\mathbf{θ} \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\left[\frac{\mathbf{1}}{\mathbf{16}}\mathbf{ln}\left(\mathbf{sec}\mathbf{4}\mathbf{θ}\mathbf{+}\mathbf{tan}\mathbf{4}\mathbf{θ}\right)\right]\mathbf{cos}\mathbf{4}\mathbf{θ} \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{16}}\mathbf{ln}\left(\mathbf{sec}\mathbf{4}\mathbf{θ}\mathbf{+}\mathbf{tan}\mathbf{4}\mathbf{θ}\right)\mathbf{cos}\mathbf{4}\mathbf{θ} \end{gathered}$$

Therefore, the general solution is,

$\begin{array}{rcl}\mathbf{y}& \mathbf{=}& {\mathbf{y}}_{\mathbf{c}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\\ & =& \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{cos}\left(\mathbf{4}\mathbf{θ}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sin}\left(\mathbf{4}\mathbf{θ}\right)\mathbf{-}\frac{\mathbf{1}}{\mathbf{16}}\mathbf{ln}\left(\mathbf{sec}\mathbf{4}\mathbf{θ}\mathbf{+}\mathbf{tan}\mathbf{4}\mathbf{θ}\right)\mathbf{cos}\mathbf{4}\mathbf{θ}\\ & & \end{array}$