91Ó°ÊÓ

The given information is,

$\begin{array}{l}(D^2+5D+4)y=\cos 2x\\ y_p=\frac{1}{10}\sin 2x\\ y=y_c+y_p=A{e}^{−x}+B{e}^{−4x}+\frac{1}{10}\sin 2x\\ (a_2{D}^2+a_{1}D+a_0)y=f\left(x\right)\end{array}$

$(a_2{D}^2+a_{1}D+a_0)y=0$

An equation is homogenous if and only if.$\mathbf{g}\left(\mathbf{x}\right)≡\mathbf{0}$

A general solution to the nth order differential equation is one that incorporates a significant number of arbitrary constants. If one use the variable approach to solve a first-order differential equation, one must insert an arbitrary constant as soon as integration is completed.

The given equation is,

$\begin{array}{c}(D^2+5D+4)y=\cos 2x\\ y_p=\frac{1}{10}\sin 2x\\ (D^2+5D+4)y=D^{2}y+5Dy+4y\\ y^{\text{'}\text{'}}+4y^{\text{'}}+5y\end{array}$

Solve the problem further

$\begin{array}{l}{y}^{\text{'}\text{'}}+4y^{\text{'}}+5y=\cos \left(2x\right)\\ y_p^{\text{'}\text{'}}+5y_p^{\text{'}}+4y_p=\cos \left(2x\right)\end{array}$

Hence,

$\begin{array}{l}{y}_p=\frac{1}{10}\sin 2x\\ y^{\text{'}}=\frac{2}{10}\cos 2x\\ y^{\text{'}\text{'}}=\frac{−4}{10}\sin 2x\end{array}$

Substitute the value and solve

$\begin{array}{c}{y}_p^{\text{'}\text{'}}+5y_p^{\text{'}}+4y_p=\cos \left(2x\right)\\ \frac{−4}{10}\sin 2x+5\left(\frac{2}{10}\cos 2x\right)+4\left(\frac{1}{10}\sin 2x\right)=\cos 2x\\ \cos 2x=\cos 2x\end{array}$

$y_p=\frac{1}{10}\sin 2x$is a solution of$(D^2+5D+4)y=\cos 2x$

$\begin{array}{l}{y}_p=\frac{1}{10}\sin 2x−e^{−x}\\ y^{\text{'}}=\frac{2}{10}\cos 2x+e^{−x}\\ y^{\text{'}\text{'}}=\frac{−4}{10}\sin 2x−e^{−x}\end{array}$

Proceed further

$\begin{array}{c}{y}_p^{\text{'}\text{'}}+5y_p^{\text{'}}+4y_p=\cos \left(2x\right)\\ (\frac{−4}{10}\sin 2x−e^{−x})+5(\frac{2}{10}\cos 2x+e^{−x})+4(\frac{1}{10}\sin 2x−e^{−x})=\cos 2x\\ \cos 2x=\cos 2x\end{array}$

$y_p=\frac{1}{10}\sin 2x−e^{−x}$is particular solution of$(D^2+5D+4)y=\cos 2x$

The difference between two particular solutions$(a_2{D}^2+a_{1}D+a_0)y$

$=f\left(x\right)$is solution of$(a_2{D}^2+a_{1}D+a_0)y=0$

Let

$y_p$and$y_c$ be particular solution of$(a_2{D}^2+a_{1}D+a_0)y=f\left(x\right)$

$\begin{array}{l}(a_2{D}^2+a_{1}D+a_0)y_p=f\left(x\right)\\ (a_2{D}^2+a_{1}D+a_0)y_c=f\left(x\right)\end{array}$

Substitute and find the solution by,

$\begin{array}{c}y=y_c−y_p\\ (a_2{D}^2+a_{1}D+a_0)(y_c−y_p)=\left[(a_2{D}^2+a_{1}D+a_0)y_c\right]−\left[(a_2{D}^2+a_{1}D+a_0)y_p\right]\\ =f\left(x\right)−f\left(x\right)\\ =0\end{array}$

$∴(y_c−y_p)$satisfies$(a_2{D}^2+a_{1}D+a_0)=0$

Thus proved