91Ó°ÊÓ

The differential equation is,

$\mathrm{y}\text{'}\text{'}+4\mathrm{y}=\mathrm{²õ¾\pm ²Ôθ}-\mathrm{³¦´Ç²õθ}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}......\left(1\right)$

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

$\mathrm{y}\text{'}\text{'}+4\mathrm{y}=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^2+4=0$

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2+4=0\\ \mathrm{m}=Â\pm 2\mathrm{i}\end{array}$

The roots of the auxiliary equation are,

${\mathrm{m}}_1=2\mathrm{i},\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}{\mathrm{m}}_2=-2\mathrm{i}$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1\cos \left(2\mathrm{θ}\right)+{\mathrm{c}}_2\sin \left(2\mathrm{θ}\right)$

According to the method of undetermined coefficients, to find a particular solution to the differential equation;

$\mathrm{ay}\text{'}\text{'}+\mathrm{by}\text{'}+\mathrm{cy}=\left\{\begin{array}{l}{\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{³¦´Ç²õβ³Ù}\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰}\mathrm{or}\\ {\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{²õ¾\pm ²Ôβ³Ù}\end{array}\right\}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰}......\left(2\right)$

For $\mathrm{β}≠0$, use the form

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{t}}^{\mathrm{s}}[({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{³¦´Ç²õβ³Ù}+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{²õ¾\pm ²Ôβ³Ù}]$

With s = 1 if$\mathrm{Î\pm }+\mathrm{¾\pm β}$ is a root of the associated auxiliary equation.

And s = 0 if $\mathrm{Î\pm }+\mathrm{¾\pm β}$is not a root of the associated auxiliary equation.

Comparing equations (1) and (2), we get:

M=0 and $\mathrm{Î\pm }=0;\mathrm{β}=1$

Therefore, $\mathbf{Î\pm }\mathbf{+}\mathbf{¾\pm β}\mathbf{=}\mathbf{i}$is not a root of the associated auxiliary equation so here s = 0

Assume, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^{\mathrm{s}}({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{Î\pm ³\ae }}\mathrm{³¦´Ç²õβ³\ae }+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{Î\pm ³\ae }}\mathrm{²õ¾\pm ²Ôβ³\ae }\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{θ}\right)={\mathrm{θ}}^0\left({\mathrm{A}}_0\right){\mathrm{e}}^{\left(0\right)\mathrm{θ}}\cos \left(1\right)\mathrm{θ}+{\mathrm{θ}}^0\left({\mathrm{B}}_0\right){\mathrm{e}}^{\left(0\right)\mathrm{θ}}\sin \left(1\right)\mathrm{θ}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{θ}\right)={\mathrm{A}}_0\mathrm{³¦´Ç²õθ}+{\mathrm{B}}_0\mathrm{²õ¾\pm ²Ôθ}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{θ}\right)=\mathrm{A³¦´Ç²õθ}+\mathrm{B²õ¾\pm ²Ôθ}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰}......\left(3\right)\end{array}$

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

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

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

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+4\mathrm{y}=\mathrm{²õ¾\pm ²Ôθ}-\mathrm{³¦´Ç²õθ}\\ -\mathrm{A³¦´Ç²õθ}-\mathrm{B²õ¾\pm ²Ôθ}+4(\mathrm{A³¦´Ç²õθ}+\mathrm{B²õ¾\pm ²Ôθ})=\mathrm{²õ¾\pm ²Ôθ}-\mathrm{³¦´Ç²õθ}\\ 3\mathrm{A³¦´Ç²õθ}+3\mathrm{B²õ¾\pm ²Ôθ}=\mathrm{²õ¾\pm ²Ôθ}-\mathrm{³¦´Ç²õθ}\end{array}$

Comparing all coefficients of the above equation,

role="math" localid="1654947115962" $\begin{array}{c}3\mathrm{A}=-1\text{ â¶Ä‰}⇒\mathrm{A}=\frac{-1}{3}\\ 3\mathrm{B}=1\text{ â¶Ä‰}⇒\mathrm{B}=\frac{1}{3}\end{array}$

Substitute the value of A and B in the equation (3),

role="math" localid="1654947233973" $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{θ}\right)=\mathrm{A³¦´Ç²õθ}+\mathrm{B²õ¾\pm ²Ôθ}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{θ}\right)=\left(-\frac{1}{3}\right)\mathrm{³¦´Ç²õθ}+\left(\frac{1}{3}\right)\mathrm{²õ¾\pm ²Ôθ}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{θ}\right)=\frac{1}{3}\mathrm{²õ¾\pm ²Ôθ}-\frac{1}{3}\mathrm{³¦´Ç²õθ}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{θ}\right)=\frac{1}{3}(\mathrm{²õ¾\pm ²Ôθ}-\mathrm{³¦´Ç²õθ})\end{array}$

Therefore, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{θ}\right)=\frac{1}{3}(\mathrm{²õ¾\pm ²Ôθ}-\mathrm{³¦´Ç²õθ})$

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{θ}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{θ}\right)\\ \mathrm{y}={\mathrm{c}}_1\cos \left(2\mathrm{θ}\right)+{\mathrm{c}}_2\sin \left(2\mathrm{θ}\right)+\frac{1}{3}(\mathrm{²õ¾\pm ²Ôθ}-\mathrm{³¦´Ç²õθ})\end{array}$