91Ó°ÊÓ

The differential equation is $y^{\text{'}\text{'}}+4y^{\text{'}}+5y=2e^{−2x}\cos x$.

When fand its derivatives are inserted into the equation, a solution is a function y = f (x) that solves the differential equation. The highest order of any derivative of the unknown function appearing in the equation is the order of a differential equation.

A differential equation of the form$\mathbf{(}\mathit{D}\mathbf{−}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{−}\mathit{b}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\text{ }}\mathit{a}\mathbf{≠}\mathit{b}$ has general solution $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{a}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{b}\mathbf{x}}\mathbf{.}$

Consider the equation.

$y^{\text{'}\text{'}}+4y^{\text{'}}+5y=2e^{−2x}\cos x$

The above equation is a non-homogenous equation.

Substitute the values in above equation.

The auxiliary equation is,

$\begin{array}{c}{m}^2+4m+5=0\\ m=−2−i,−2+i\end{array}$

The solution for$m=−2−i,−2+i$ is,

$y_c=e^{−2x}\left(c_1\cos x+c_2\sin x\right)$

The solution of equation is,

$y=y_c+y_p…………….\left(1\right)$

The value of y_pis,

$\begin{array}{c}{y}_p=\frac{2e^{(−2+i)x}}{\left(D^2+4D+5\right)}\\ =\frac{2e^{(−2+i)x}}{{(D−2+i)}^2+4(D−2+i)(+5)}\\ =x{e}^{−2x}\sin x\end{array}$

Substitute the values of y_p and y_c in Equation (1).

$y=e^{−2x}\left(c_1\cos x+c_2\sin x\right)+x{e}^{−2x}\sin x$

Thus, the solution of given differential equation is

$y=e^{−2x}\left(c_1\cos x+c_2\sin x\right)+x{e}^{−2x}\sin x$