Find the solution to the given initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{+}\mathbf{y}\left(\mathbf{\theta }\right)\mathbf{=}\mathbf{2}\mathbf{cos\theta }\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{3}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}$

The differential equation is,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{+}\mathbf{y}\left(\mathbf{\theta }\right)\mathbf{=}\mathbf{2}\mathbf{cos\theta } ......\left(\mathbf{1}\right)$

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

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

The auxiliary equation for the above equation,

$\begin{array}{rcl}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{1}& \mathbf{=}& \mathbf{0}\\ {\left(\mathbf{m}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}& \mathbf{=}& \mathbf{0}\\ & & \end{array}$

The root of an auxiliary equation is ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}$

The complementary solution of the given equation is${\mathbf{y}}_{\mathbf{c}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{\theta }}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{\theta e}}^{\mathbf{-}\mathbf{\theta }}$

Assume, the particular solution of equation (1),

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

Now find the first and second derivatives of the above equation,

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{=}\mathbf{-}\mathbf{Asin\theta }\mathbf{+}\mathbf{Bcos\theta } \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{=}\mathbf{-}\mathbf{Acos\theta }\mathbf{-}\mathbf{Bsin\theta } \end{gathered}$$

Substitute the value of ${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{\theta }\right)\mathbf{,} {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\left(\mathbf{\theta }\right)$and ${\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{\theta }\right)$the equation (1),

$\begin{array}{rcl}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{+}\mathbf{y}\left(\mathbf{\theta }\right)& \mathbf{=}& \mathbf{2}\mathbf{cos\theta }\\ \mathbf{-}\mathbf{Acos\theta }\mathbf{-}\mathbf{Bsin\theta }\mathbf{+}\mathbf{2}\left[\mathbf{-}\mathbf{Asin\theta }\mathbf{+}\mathbf{Bcos\theta }\right]\mathbf{+}\mathbf{Acos\theta }\mathbf{+}\mathbf{Bsin\theta }& \mathbf{=}& \mathbf{2}\mathbf{cos\theta }\\ \mathbf{-}\mathbf{2}\mathbf{Asin\theta }\mathbf{+}\mathbf{2}\mathbf{Bcos\theta }& \mathbf{=}& \mathbf{2}\mathbf{cos\theta }\\ & & \end{array}$

Comparing all coefficients of the above equation,

$$\begin{gathered} \mathbf{-}\mathbf{2}\mathbf{A}\mathbf{=}\mathbf{0} \Rightarrow \mathbf{A}\mathbf{=}\mathbf{0} \\ \mathbf{2}\mathbf{B}\mathbf{=}\mathbf{2} \Rightarrow \mathbf{B}\mathbf{=}\mathbf{1} \end{gathered}$$

Substitute the value of A, B and C in the equation (2),

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{\theta }\right)\mathbf{=}\mathbf{Acos\theta }\mathbf{+}\mathbf{Bsin\theta } \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{\theta }\right)\mathbf{=}\left(\mathbf{0}\right)\mathbf{cos\theta }\mathbf{+}\left(\mathbf{1}\right)\mathbf{sin\theta } \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{\theta }\right)\mathbf{=}\mathbf{sin\theta } \end{gathered}$$

Therefore, the general solution is,

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{c}}\left(\mathbf{\theta }\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{\theta }\right) \\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{\theta }}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{\theta e}}^{\mathbf{-}\mathbf{\theta }}\mathbf{+}\mathbf{sin\theta } ......\left(\mathbf{3}\right) \end{gathered}$$

Given initial condition,

$\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{3}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}$

Substitute the value of $y=3$and $\theta =0$in the equation (3),

$$\begin{gathered} \mathbf{3}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\left(\mathbf{0}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\left(\mathbf{0}\right){\mathbf{e}}^{\mathbf{-}\left(\mathbf{0}\right)}\mathbf{+}\mathbf{sin}\left(\mathbf{0}\right) \\ {\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{3} \end{gathered}$$

Now find the derivative of the equation (3),

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{\theta }}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{\theta }}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}{\mathbf{\theta e}}^{\mathbf{-}\mathbf{\theta }}\mathbf{+}\mathbf{cos\theta }$

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

$\begin{array}{rcl}\mathbf{0}& \mathbf{=}& \mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\left(\mathbf{0}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\left(\mathbf{0}\right)}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}\left(\mathbf{0}\right){\mathbf{e}}^{\mathbf{-}\left(\mathbf{0}\right)}\mathbf{+}\mathbf{cos}\left(\mathbf{0}\right)\\ \mathbf{0}& \mathbf{=}& \mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{+}\mathbf{1}\\ {\mathbf{c}}_{\mathbf{1}}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}& \mathbf{=}& \mathbf{1} ......\left(\mathbf{4}\right)\\ & & \end{array}$

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

$\begin{array}{rcl}{\mathbf{c}}_{\mathbf{1}}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}& \mathbf{=}& \mathbf{1}\\ \mathbf{3}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}& \mathbf{=}& \mathbf{1}\\ {\mathbf{c}}_{\mathbf{2}}& \mathbf{=}& \mathbf{2}\\ & & \end{array}$

Substitute the value of $c_1$and $c_2$in the equation (3),

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{\theta }}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{\theta e}}^{\mathbf{-}\mathbf{\theta }}\mathbf{+}\mathbf{sin\theta } \\ \mathbf{y}\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{\theta }}\mathbf{+}\mathbf{2}{\mathbf{\theta e}}^{\mathbf{-}\mathbf{\theta }}\mathbf{+}\mathbf{sin\theta } \end{gathered}$$

Therefore, the general solution is$\mathbf{y}\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{\theta }}\mathbf{+}\mathbf{2}{\mathbf{\theta e}}^{\mathbf{-}\mathbf{\theta }}\mathbf{+}\mathbf{sin\theta }$.