Q23RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given differential equation.

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

Step-by-Step Solution

Answer

The general solution to the given differential equation is:

$\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{cos}\left(\mathbf{4}\mathbf{\theta }\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sin}\left(\mathbf{4}\mathbf{\theta }\right)\mathbf{-}\frac{\mathbf{1}}{\mathbf{16}}\mathbf{ln}\left(\mathbf{sec}\mathbf{4}\mathbf{\theta }\mathbf{+}\mathbf{tan}\mathbf{4}\mathbf{\theta }\right)\mathbf{cos}\mathbf{4}\mathbf{\theta }$

Write the auxiliary equation of the given differential equation

The given differential equation is,

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

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

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

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

Find the roots of the auxiliary equation

Solve the auxiliary equation,

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

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 ${\mathbf{y}}_{\mathbf{c}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{cos}\left(\mathbf{4}\mathbf{\theta }\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sin}\left(\mathbf{4}\mathbf{\theta }\right)$.

Find the particular solution

Assume, the particular solution of equation (1),

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

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

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

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

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

Therefore, the particular solution of equation (1),

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

Conclusion, write the general solution

Therefore, the general solution is,

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