91Ó°ÊÓ

The given differential equation is:

$\mathrm{y}\text{'}\text{'}+4\mathrm{y}=8\sin \left(2\mathrm{t}\right)\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}…\left(1\right)$

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:

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

The complementary solution of the given equation is;

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1\cos 2\mathrm{t}+{\mathrm{c}}_2\sin 2\mathrm{t}$

According to the method of undetermined coefficients, assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}=\mathrm{t}[\mathrm{Asin}\left(2\mathrm{t}\right)+\mathrm{Bcos}\left(2\mathrm{t}\right)]\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰}......\left(2\right)$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}=\mathrm{t}[\mathrm{Asin}\left(2\mathrm{t}\right)+\mathrm{Bcos}\left(2\mathrm{t}\right)]\\ {\mathrm{y}}_{\mathrm{p}}\text{'}=(2\mathrm{At}+\mathrm{B})\cos \left(2\mathrm{t}\right)+(\mathrm{A}-2\mathrm{Bt})\sin \left(2\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=(4\mathrm{A}-4\mathrm{Bt})\cos \left(2\mathrm{t}\right)+(-4\mathrm{B}-4\mathrm{At})\sin \left(2\mathrm{t}\right)\end{array}$

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}+4{\mathrm{y}}_{\mathrm{p}}=8\sin \left(2\mathrm{t}\right)\\ (4\mathrm{A}-4\mathrm{Bt})\cos \left(2\mathrm{t}\right)+(-4\mathrm{B}-4\mathrm{At})\sin \left(2\mathrm{t}\right)+4\left(\mathrm{t}[\mathrm{Asin}\left(2\mathrm{t}\right)+\mathrm{Bcos}\left(2\mathrm{t}\right)]\right)=8\sin \left(2\mathrm{t}\right)\\ 4\mathrm{Acos}\left(2\mathrm{t}\right)-4\mathrm{Bsin}\left(2\mathrm{t}\right)=8\sin \left(2\mathrm{t}\right)\end{array}$

Comparing all coefficients of the above equation;

$\begin{array}{c}4\mathrm{A}=0\text{ â¶Ä‰}⇒\mathrm{A}=0\\ -4\mathrm{B}=8\text{ â¶Ä‰}⇒\mathrm{B}=-2\end{array}$

Therefore, the particular solution of equation (1),

$\begin{array}{l}{\mathrm{y}}_{\mathrm{p}}=\mathrm{t}[\mathrm{Asin}\left(2\mathrm{t}\right)+\mathrm{Bcos}\left(2\mathrm{t}\right)]\\ {\mathrm{y}}_{\mathrm{p}}=-2\mathrm{tcos}\left(2\mathrm{t}\right)\end{array}$