91Ó°ÊÓ

Given equation is$y^{\text{'}\text{'}}+4y^{\text{'}}+8y=30e^{−x/2}\cos 5x/2$

Concept Used:

$\mathbf{\{}\begin{array}{l}\mathbf{C}{\mathbf{e}}^{\mathbf{c}\mathbf{x}}\mathbf{\text{if}}\mathbf{c}\mathbf{\text{isnotequaltoeithera}}\\ \mathbf{C}\mathbf{x}{\mathbf{e}}^{\mathbf{c}\mathbf{x}}\mathbf{\text{if}}\mathbf{c}\mathbf{\text{equalsaor}}\mathbf{b}\mathbf{,}\mathbf{a}\mathbf{≠}\mathbf{b}\\ \mathbf{C}{\mathbf{x}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{c}\mathbf{x}}\mathbf{\text{if}}\mathbf{c}\mathbf{=}\mathbf{a}\mathbf{=}\mathbf{b}\end{array}$

$\mathbf{(}\mathit{D}\mathbf{−}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{−}\mathit{b}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{\{}\begin{array}{l}\mathbf{k}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{Î\pm }\mathbf{x}\\ \mathbf{k}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{Î\pm }\mathbf{x}\end{array}$

First solve

$\begin{array}{l}\mathbf{(}\mathbf{D}\mathbf{−}\mathbf{a}\mathbf{)}\mathbf{(}\mathbf{D}\mathbf{−}\mathbf{b}\mathbf{)}\mathbf{y}\mathbf{=}\mathbf{k}{\mathbf{e}}^{\mathbf{i}\mathbf{Î\pm }\mathbf{x}}\\ \mathbf{(}\mathbf{D}\mathbf{−}\mathbf{a}\mathbf{)}\mathbf{(}\mathbf{D}\mathbf{−}\mathbf{b}\mathbf{)}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{c}\mathbf{x}}{\mathbf{P}}_{\mathbf{n}}\\ {\mathbf{y}}_{\mathbf{p}}\mathbf{=}\mathbf{\{}\begin{array}{l}{\mathbf{e}}^{\mathbf{c}\mathbf{x}}{\mathbf{Q}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\text{if}}\mathbf{c}\mathbf{≠}\mathbf{a}\mathbf{\text{or}}\mathbf{b}\\ \mathbf{x}{\mathbf{e}}^{\mathbf{c}\mathbf{x}}{\mathbf{Q}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\text{if}}\mathbf{c}\mathbf{=}\mathbf{a}\mathbf{\text{or}}\mathbf{b}\mathbf{\text{but}}\mathbf{a}\mathbf{≠}\mathbf{b}\\ {\mathbf{x}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{c}\mathbf{x}}{\mathbf{Q}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\text{if}}\mathbf{c}\mathbf{=}\mathbf{a}\mathbf{=}\mathbf{b}\end{array}\end{array}$

On solving differential equation

$\begin{array}{c}{D}^2+4D+8y=30e^{(−1+5i)x/2}\\ (D^2+4D+8)y=30e^{(−1+5i)x/2}\\ (D+2−2i)(D+2+2i)y=30e^{(−1+5i)x/2}\\ (D+2+2i)(D+2−2i)y=0\end{array}$

Hence the result is,

$y_c=A{e}^{−2x}\sin (2x+\mathrm{γ})$

Let

$\begin{array}{c}u=(D+2+2i)y\\ (D+2−2i)u=30e^{(−1+5i)x/2}\\ u^{\text{'}}+(D+2−2i)u=30e^{(−1+5i)x/2}\end{array}$

Solve the equation further

$\begin{array}{l}⇒e^I=e^{(2−2i)x}\\ u{e}^I=∫\left(30e^{(−1+5i)x/2}\right)e^{(2−2i)x}dx\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}=(18−6i)e^{(3+i)x/2}\\ u\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}=(18−6i)e^{(−1+5i)x/2}\end{array}$

On substituting in

$\begin{array}{l}u=(D+2+2i)y\\ ⇒(18−6i)e^{(−1+5i)x/2}\\ =(D+2+2i)y\\ ⇒(18−6i)e^{(−1+5i)x/2}\\ =y^{\text{'}}+(2+2i)y\end{array}$

Again,

$\begin{array}{c}I=∫(2+2i)dx\\ =(2+2i)x\end{array}$

Solve the equation further

$⇒e^I=e^{(2+2i)x}$

Substitute and solve further

$\begin{array}{c}{y}_p{e}^I=∫\left[(18−6i)e^{(−1+5i)x/2}\right]e^{(2+2i)x}dx\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â€‰}=−4i{e}^{(3+9i)x/2}\\ y_p\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}=−4i{e}^{(−1+5i)x/2}\end{array}$

Therefore, the general solution of

$\begin{array}{l}(D^2+4D+12)y=80\sin 2x\text{is}\\ \mathrm{Re}\left[y_p\right]=4e^{−x/2}\sin \frac{5}{2}x\\ y=y_c+\mathrm{Re}\left[y_p\right]\end{array}$

That is,

$⇒e^I=e^{(2+2i)x}$