91Ó°ÊÓ

The differential equation is,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{θ}\right)\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{θ}\right)\mathbf{+}\mathbf{y}\left(\mathbf{θ}\right)\mathbf{=}\mathbf{2}\mathbf{³¦´Ç²õθ}                     ......\left(\mathbf{1}\right)$

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

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

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${\mathbf{y}}_{\mathbf{c}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{θ}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{θ\pm ð}}^{\mathbf{-}\mathbf{θ}}$

Assume, the particular solution of equation (1),

${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{θ}\right)\mathbf{=}\mathbf{A³¦´Ç²õθ}\mathbf{+}\mathbf{B²õ¾\pm ²Ôθ}               ......\left(\mathbf{2}\right)$

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

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\left(\mathbf{θ}\right)\mathbf{=}\mathbf{-}\mathbf{A²õ¾\pm ²Ôθ}\mathbf{+}\mathbf{B³¦´Ç²õθ} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{θ}\right)\mathbf{=}\mathbf{-}\mathbf{A³¦´Ç²õθ}\mathbf{-}\mathbf{B²õ¾\pm ²Ôθ} \end{gathered}$$

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

$\begin{array}{rcl}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{θ}\right)\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{θ}\right)\mathbf{+}\mathbf{y}\left(\mathbf{θ}\right)& \mathbf{=}& \mathbf{2}\mathbf{³¦´Ç²õθ}\\ \mathbf{-}\mathbf{A³¦´Ç²õθ}\mathbf{-}\mathbf{B²õ¾\pm ²Ôθ}\mathbf{+}\mathbf{2}\left[\mathbf{-}\mathbf{A²õ¾\pm ²Ôθ}\mathbf{+}\mathbf{B³¦´Ç²õθ}\right]\mathbf{+}\mathbf{A³¦´Ç²õθ}\mathbf{+}\mathbf{B²õ¾\pm ²Ôθ}& \mathbf{=}& \mathbf{2}\mathbf{³¦´Ç²õθ}\\ \mathbf{-}\mathbf{2}\mathbf{A²õ¾\pm ²Ôθ}\mathbf{+}\mathbf{2}\mathbf{B³¦´Ç²õθ}& \mathbf{=}& \mathbf{2}\mathbf{³¦´Ç²õθ}\\ & & \end{array}$

Comparing all coefficients of the above equation,

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

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

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{θ}\right)\mathbf{=}\mathbf{A³¦´Ç²õθ}\mathbf{+}\mathbf{B²õ¾\pm ²Ôθ} \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{θ}\right)\mathbf{=}\left(\mathbf{0}\right)\mathbf{³¦´Ç²õθ}\mathbf{+}\left(\mathbf{1}\right)\mathbf{²õ¾\pm ²Ôθ} \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{θ}\right)\mathbf{=}\mathbf{²õ¾\pm ²Ôθ} \end{gathered}$$

Therefore, the general solution is,

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{c}}\left(\mathbf{θ}\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{θ}\right) \\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{θ}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{θ\pm ð}}^{\mathbf{-}\mathbf{θ}}\mathbf{+}\mathbf{²õ¾\pm ²Ôθ}                     ......\left(\mathbf{3}\right) \end{gathered}$$

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 $\mathrm{θ}=0$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),

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{θ}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{θ}}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}{\mathbf{θ\pm ð}}^{\mathbf{-}\mathbf{θ}}\mathbf{+}\mathbf{³¦´Ç²õθ}$

Substitute the value of $y^{\text{'}}=0$and $\mathrm{θ}=0$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),

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{θ}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{θ\pm ð}}^{\mathbf{-}\mathbf{θ}}\mathbf{+}\mathbf{²õ¾\pm ²Ôθ} \\ \mathbf{y}\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{θ}}\mathbf{+}\mathbf{2}{\mathbf{θ\pm ð}}^{\mathbf{-}\mathbf{θ}}\mathbf{+}\mathbf{²õ¾\pm ²Ôθ} \end{gathered}$$

Therefore, the general solution is$\mathbf{y}\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{θ}}\mathbf{+}\mathbf{2}{\mathbf{θ\pm ð}}^{\mathbf{-}\mathbf{θ}}\mathbf{+}\mathbf{²õ¾\pm ²Ôθ}$.