91Ó°ÊÓ

The given differential equation is,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{θ}\right)\mathbf{+}\mathbf{y}\left(\mathbf{θ}\right)\mathbf{=}\mathbf{²õ\pm ð³¦Î¸}                     ......\left(\mathbf{1}\right)$

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

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

The auxiliary equation for the above equation,

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

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,

${\mathbf{y}}_{\mathbf{c}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{³¦´Ç²õθ}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{²õ¾\pm ²Ôθ}$

Assume, the particular solution of equation (1),

${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{θ}\right)\mathbf{=}{\mathbf{v}}_{\mathbf{1}}\mathbf{³¦´Ç²õθ}\mathbf{+}{\mathbf{v}}_{\mathbf{2}}\mathbf{²õ¾\pm ²Ôθ}               ......\left(\mathbf{2}\right)$

Now,

$\begin{array}{rcl}{\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}\mathbf{³¦´Ç²õθ}\mathbf{+}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{²õ¾\pm ²Ôθ}& \mathbf{=}& \mathbf{0}\\ {\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}\mathbf{³¦´Ç²õθ}& \mathbf{=}& \mathbf{-}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{²õ¾\pm ²Ôθ}\\ {\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}& \mathbf{=}& \mathbf{-}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\frac{\mathbf{²õ¾\pm ²Ôθ}}{\mathbf{³¦´Ç²õθ}}\\ {\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}& \mathbf{=}& \mathbf{-}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{³Ù²¹²Ôθ}                    ......\left(\mathbf{3}\right)\\ & & \end{array}$

Also,

$\mathbf{-}{\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}\mathbf{²õ¾\pm ²Ôθ}\mathbf{+}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{³¦´Ç²õθ}\mathbf{=}\mathbf{²õ\pm ð³¦Î¸}$

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

$\begin{array}{rcl}\mathbf{-}\left(\mathbf{-}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{³Ù²¹²Ôθ}\right)\mathbf{²õ¾\pm ²Ôθ}\mathbf{+}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{³¦´Ç²õθ}& \mathbf{=}& \mathbf{²õ\pm ð³¦Î¸}\\ {\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\frac{{\mathbf{sin}}^{\mathbf{2}}\mathbf{θ}}{\mathbf{³¦´Ç²õθ}}\mathbf{+}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{³¦´Ç²õθ}& \mathbf{=}& \mathbf{²õ\pm ð³¦Î¸}\\ \frac{{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}{\mathbf{sin}}^{\mathbf{2}}\mathbf{θ}\mathbf{+}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}{\mathbf{cos}}^{\mathbf{2}}\mathbf{θ}}{\mathbf{³¦´Ç²õθ}}& \mathbf{=}& \mathbf{²õ\pm ð³¦Î¸}\\ \frac{{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}}{\mathbf{³¦´Ç²õθ}}& \mathbf{=}& \mathbf{²õ\pm ð³¦Î¸}\\ {\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}& \mathbf{=}& \mathbf{1}\\ {\mathbf{v}}_{\mathbf{2}}& \mathbf{=}& \mathbf{θ}\end{array}$

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

$$\begin{gathered} {\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}\mathbf{=}\mathbf{-}{\mathbf{v}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{³Ù²¹²Ôθ} \\ {\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\left(\mathbf{1}\right)\mathbf{³Ù²¹²Ôθ} \\ {\mathbf{v}}_{\mathbf{1}}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{³Ù²¹²Ôθ} \\ {\mathbf{v}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{ln}\left|\mathbf{²õ\pm ð³¦Î¸}\right| \\ {\mathbf{v}}_{\mathbf{1}}\mathbf{=}\mathbf{ln}\left|\mathbf{³¦´Ç²õθ}\right| \end{gathered}$$

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

${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{θ}\right)\mathbf{=}\left[\mathbf{ln}\left|\mathbf{³¦´Ç²õθ}\right|\right]\mathbf{³¦´Ç²õθ}\mathbf{+}\mathbf{θ²õ¾\pm ²Ôθ}$

Therefore, the general solution is,

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{c}}\left(\mathbf{θ}\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{θ}\right) \\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{³¦´Ç²õθ}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{²õ¾\pm ²Ôθ}\mathbf{+}\left[\mathbf{ln}\left|\mathbf{³¦´Ç²õθ}\right|\right]\mathbf{³¦´Ç²õθ}\mathbf{+}\mathbf{θ²õ¾\pm ²Ôθ}                    ......\left(\mathbf{4}\right) \end{gathered}$$

The given initial condition are,

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

Substitute the value of $y=1$and $\mathrm{θ}=0$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),

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{²õ¾\pm ²Ôθ}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{³¦´Ç²õθ}\mathbf{+}\left(\mathbf{-}\mathbf{²õ¾\pm ²Ôθ}\right)\mathbf{+}\left[\mathbf{ln}\left|\mathbf{³¦´Ç²õθ}\right|\right]\left(\mathbf{-}\mathbf{²õ¾\pm ²Ôθ}\right)\mathbf{+}\mathbf{賦´Ç²õθ}\mathbf{+}\mathbf{²õ¾\pm ²Ôθ}$

Substitute the value of $y^{\text{'}}=2$and $\mathrm{θ}=0$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),

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{³¦´Ç²õθ}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{²õ¾\pm ²Ôθ}\mathbf{+}\left[\mathbf{ln}\left|\mathbf{³¦´Ç²õθ}\right|\right]\mathbf{³¦´Ç²õθ}\mathbf{+}\mathbf{θ²õ¾\pm ²Ôθ} \\ \mathbf{y}\mathbf{=}\left(\mathbf{1}\right)\mathbf{³¦´Ç²õθ}\mathbf{+}\left(\mathbf{2}\right)\mathbf{²õ¾\pm ²Ôθ}\mathbf{+}\left[\mathbf{ln}\left|\mathbf{³¦´Ç²õθ}\right|\right]\mathbf{³¦´Ç²õθ}\mathbf{+}\mathbf{θ²õ¾\pm ²Ôθ} \\ \mathbf{y}\mathbf{=}\mathbf{³¦´Ç²õθ}\mathbf{+}\mathbf{2}\mathbf{²õ¾\pm ²Ôθ}\mathbf{+}\left[\mathbf{ln}\left|\mathbf{³¦´Ç²õθ}\right|\right]\mathbf{³¦´Ç²õθ}\mathbf{+}\mathbf{θ²õ¾\pm ²Ôθ} \end{gathered}$$

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

$\mathbf{y}\mathbf{=}\mathbf{³¦´Ç²õθ}\mathbf{+}\mathbf{2}\mathbf{²õ¾\pm ²Ôθ}\mathbf{+}\left[\mathbf{ln}\left|\mathbf{³¦´Ç²õθ}\right|\right]\mathbf{³¦´Ç²õθ}\mathbf{+}\mathbf{θ²õ¾\pm ²Ôθ}$