Question

: Find the general solution of the following differential equations (complementary function + particular solution). Find the particular solution by inspection or by$\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{18}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{23}\mathbf{)}\mathbf{,}$or.$\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{24}\mathbf{)}\mathbf{,}$Also find a computer solution and reconcile differences if necessary, noticing especially whether the particular solution is in simplest form [see$\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{26}\mathbf{)}\mathbf{,}$and the discussionafter$\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{15}\mathbf{)}\mathbf{,}$].

$\mathbf{(}\mathbf{4}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{5}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{40}{\mathit{e}}^{\mathbf{-}\mathbf{3}\mathbf{x}\mathbf{/}\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{x}$

Short answer

The general solution given by differential equation is

$e^{-x/2}(A\sin x+B\cos x)+e^{-3x/2}(-4\sin 2x+8\cos 2x)$

Step-by-step solution

Given data.

Given equation is$(4D^2+4D+5)y=40e^{-3x/2}\sin 2x$

General solution of differential equation. 

Concept Used:

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

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

first solve

width="313">(D−a)(D−b)y=keiαx(D−a)(D−b)y=ecxPnyp={ecxQn(x)ifc≠aorbxecxQn(x)ifc=aorbbuta≠bx2ecxQn(x)ifc=a=b

Find the general solution of given differential equation.(4D2+4D+5)y=40e−3x/2sin2x 

On solving differential equation$(4D^2+4D+5)y=40e^{-3x/2}\sin 2x$

$\begin{array}{c}(4D^2+4D+12)y=40e^{(-\frac{3}{2}+2i)x}\\ (D+\frac{1}{2}+i)(D+\frac{1}{2}-i)y=40e^{(-\frac{3}{2}+2i)x}\\ (D+\frac{1}{2}+i)(D+\frac{1}{2}-i)y=0\\ y_c=e^{-x/2}(A\sin x+B\cos x)\end{array}$

Let

$\begin{array}{c}u=(D+\frac{1}{2}-i)y\\ (D+\frac{1}{2}-i)u=40e^{(-\frac{3}{2}+2i)x}\\ u^{\text{'}}+(D+\frac{1}{2}-i)u=40e^{(-\frac{3}{2}+2i)x}\end{array}$

Solve the equation further

$\begin{array}{c}I=\int (\frac{1}{2}+i)dx\\ =(\frac{1}{2}+i)x\\ e^I=e^{(\frac{1}{2}+i)x}\end{array}$

Substitute the value in the equation

$\begin{array}{c}u{e}^I=\int \left(40e^{(-\frac{3}{2}+2i)x}\right)e^{(\frac{1}{2}+i)x}dx\\ =(-4-12i)e^{(-1+3i)x}\\ u=(-4-12i)e^{(-\frac{3}{2}+2i)x}\end{array}$

On substituting in

$\begin{array}{c}u=(D+\frac{1}{2}-i)y\\ =(-4-12i)e^{(-\frac{3}{2}+2i)x}\\ =y^{\text{'}}+(\frac{1}{2}-i)y\end{array}$

Again,

$\begin{array}{c}I=\int (\frac{1}{2}-i)dx\\ =(\frac{1}{2}-i)x\\ e^I=e^{(\frac{1}{2}-i)x}\end{array}$

Substitute the value in the equation and obtain the result

Therefore, the general solution of

$\begin{array}{l}(4D^2+4D+5)y=40e^{-3x/2}\sin 2x\text{is}\\ \mathrm{Im}\left[y_p\right]=e^{-3x/2}(-4\sin 2x+8\cos 2x)\\ y=y_c+\mathrm{Im}\left[y_p\right]\\ \text{i.e.}\\ y=e^{-x/2}(A\sin x+B\cos x)+e^{-3x/2}(-4\sin 2x+8\cos 2x)\end{array}$