Q35RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find the solution to the given initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\theta \right)\mathbf{+}\mathbf{y}\left(\theta \right)\mathbf{=}\mathbf{sec\theta }\mathbf{;}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{2}$

Step-by-Step Solution

Answer

The solution to the given initial value problem is;

$\mathbf{y}\mathbf{=}\mathbf{cos\theta }\mathbf{+}\mathbf{2}\mathbf{sin\theta }\mathbf{+}\left[\mathbf{ln}\left|\mathbf{cos\theta }\right|\right]\mathbf{cos\theta }\mathbf{+}\mathbf{\theta sin\theta }$

Write the auxiliary equation of the given differential equation

The given differential equation is,

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

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

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

The auxiliary equation for the above equation,

$\begin{array}{rcl}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{1}& \mathbf{=}& \mathbf{0}\\ {\mathbf{m}}^{\mathbf{2}}& \mathbf{=}& \mathbf{-}\mathbf{1}\\ \mathbf{m}& \mathbf{=}& \mathbf{\pm }\mathbf{i}\end{array}$

Now find the complementary solution of the given equation is,

The root of an auxiliary equation is, 

${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{i}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{i}$

The complementary solution of the given equation is,

${\mathbf{y}}_{\mathbf{c}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{cos\theta }\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sin\theta }$

Find the particular solution

Assume, the particular solution of equation (1),

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

Now,

$\begin{array}{rcl}{\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}\mathbf{cos\theta }\mathbf{+}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{sin\theta }& \mathbf{=}& \mathbf{0}\\ {\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}\mathbf{cos\theta }& \mathbf{=}& \mathbf{-}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{sin\theta }\\ {\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}& \mathbf{=}& \mathbf{-}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\frac{\mathbf{sin\theta }}{\mathbf{cos\theta }}\\ {\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}& \mathbf{=}& \mathbf{-}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{tan\theta } ......\left(\mathbf{3}\right)\\ & & \end{array}$

Also, 

 $\mathbf{-}{\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}\mathbf{sin\theta }\mathbf{+}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{cos\theta }\mathbf{=}\mathbf{sec\theta }$

Substitute the value of ${\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}$ the above equation,

$\begin{array}{rcl}\mathbf{-}\left(\mathbf{-}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{tan\theta }\right)\mathbf{sin\theta }\mathbf{+}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{cos\theta }& \mathbf{=}& \mathbf{sec\theta }\\ {\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\frac{{\mathbf{sin}}^{\mathbf{2}}\mathbf{\theta }}{\mathbf{cos\theta }}\mathbf{+}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{cos\theta }& \mathbf{=}& \mathbf{sec\theta }\\ \frac{{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\theta }\mathbf{+}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}{\mathbf{cos}}^{\mathbf{2}}\mathbf{\theta }}{\mathbf{cos\theta }}& \mathbf{=}& \mathbf{sec\theta }\\ \frac{{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}}{\mathbf{cos\theta }}& \mathbf{=}& \mathbf{sec\theta }\\ {\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}& \mathbf{=}& \mathbf{1}\\ {\mathbf{v}}_{\mathbf{2}}& \mathbf{=}& \mathbf{\theta }\end{array}$

Substitute the value of ${\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}$ in the equation (3),

$$\begin{gathered} {\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}\mathbf{=}\mathbf{-}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{tan\theta } \\ {\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\left(\mathbf{1}\right)\mathbf{tan\theta } \\ {\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{tan\theta } \\ {\mathbf{v}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{ln}\left|\mathbf{sec\theta }\right| \\ {\mathbf{v}}_{\mathbf{1}}\mathbf{=}\mathbf{ln}\left|\mathbf{cos\theta }\right| \end{gathered}$$

Substitute the value of $v_1$ and $v_2$ in the equation (2),

${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{\theta }\right)\mathbf{=}\left[\mathbf{ln}\left|\mathbf{cos\theta }\right|\right]\mathbf{cos\theta }\mathbf{+}\mathbf{\theta sin\theta }$

Find the general solution and use the given initial condition,

Therefore, the general solution is,

 $$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{c}}\left(\mathbf{\theta }\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{\theta }\right) \\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{cos\theta }\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sin\theta }\mathbf{+}\left[\mathbf{ln}\left|\mathbf{cos\theta }\right|\right]\mathbf{cos\theta }\mathbf{+}\mathbf{\theta sin\theta } ......\left(\mathbf{4}\right) \end{gathered}$$

The given initial condition are,

$\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{2}$

Substitute the value of $y=1$ and $\theta =0$ in the equation (4),

$\begin{array}{rcl}\mathbf{1}& \mathbf{=}& {\mathbf{c}}_{\mathbf{1}}\mathbf{cos}\left(\mathbf{0}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sin}\left(\mathbf{0}\right)\mathbf{+}\left[\mathbf{ln}\left|\mathbf{cos}\left(\mathbf{0}\right)\right|\right]\mathbf{cos}\left(\mathbf{0}\right)\mathbf{+}\left(\mathbf{0}\right)\mathbf{sin}\left(\mathbf{0}\right)\\ {\mathbf{c}}_{\mathbf{1}}& \mathbf{=}& \mathbf{1}\end{array}$

Now find the derivative of the equation (4),

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{sin\theta }\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{cos\theta }\mathbf{+}\left(\mathbf{-}\mathbf{sin\theta }\right)\mathbf{+}\left[\mathbf{ln}\left|\mathbf{cos\theta }\right|\right]\left(\mathbf{-}\mathbf{sin\theta }\right)\mathbf{+}\mathbf{\theta cos\theta }\mathbf{+}\mathbf{sin\theta }$

Substitute the value of $y^{\text{'}}=2$ and $\theta =0$ in the above equation,

$\begin{array}{rcl}\mathbf{2}& \mathbf{=}& \mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{sin}\left(\mathbf{0}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{cos}\left(\mathbf{0}\right)\mathbf{+}\left(\mathbf{-}\mathbf{sin}\left(\mathbf{0}\right)\right)\mathbf{+}\left[\mathbf{ln}\left|\mathbf{cos}\left(\mathbf{0}\right)\right|\right]\left(\mathbf{-}\mathbf{sin}\left(\mathbf{0}\right)\right)\mathbf{+}\left(\mathbf{0}\right)\mathbf{cos}\left(\mathbf{0}\right)\mathbf{+}\mathbf{sin}\left(\mathbf{0}\right)\\ {\mathbf{c}}_{\mathbf{2}}& \mathbf{=}& \mathbf{2}\end{array}$

Substitute the value of $c_1$and $c_2$in the equation (4),

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{cos\theta }\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sin\theta }\mathbf{+}\left[\mathbf{ln}\left|\mathbf{cos\theta }\right|\right]\mathbf{cos\theta }\mathbf{+}\mathbf{\theta sin\theta } \\ \mathbf{y}\mathbf{=}\left(\mathbf{1}\right)\mathbf{cos\theta }\mathbf{+}\left(\mathbf{2}\right)\mathbf{sin\theta }\mathbf{+}\left[\mathbf{ln}\left|\mathbf{cos\theta }\right|\right]\mathbf{cos\theta }\mathbf{+}\mathbf{\theta sin\theta } \\ \mathbf{y}\mathbf{=}\mathbf{cos\theta }\mathbf{+}\mathbf{2}\mathbf{sin\theta }\mathbf{+}\left[\mathbf{ln}\left|\mathbf{cos\theta }\right|\right]\mathbf{cos\theta }\mathbf{+}\mathbf{\theta sin\theta } \end{gathered}$$

Thus, the solution to the given initial value problem is;

$\mathbf{y}\mathbf{=}\mathbf{cos\theta }\mathbf{+}\mathbf{2}\mathbf{sin\theta }\mathbf{+}\left[\mathbf{ln}\left|\mathbf{cos\theta }\right|\right]\mathbf{cos\theta }\mathbf{+}\mathbf{\theta sin\theta }$