Question: Identify each of the differential equations in Problems 1to 24 as to type (for example, separable, linear first order, linear second order, etc.), and then solve it.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{4}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{26}{\mathit{e}}^{\mathbf{3}\mathbf{x}}$

The differential equation is $y^{\text{'}\text{'}}+4y^{\text{'}}+5y=26e^{3x}$.

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$\mathbf{(}\mathit{D}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{b}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{a}\mathbf{\ne }\mathit{b}$ 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.

$y^{\text{'}\text{'}}+4y^{\text{'}}+5y=26e^{3x}$

The above equation is a non-homogenous equation.

Substitute the values in above equation.

$\left(D^2+4D+5\right)y=26e^{3x}$

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,

$y=y_c+y_p\dots \dots \dots \dots \dots \left(1\right)$

The value of y_p is,

$\begin{array}{c}{y}_p=\frac{26}{\left(D^2+4D+5\right)}{e}^{3x}\\ =\frac{26}{\left(3^2+4\left(3\right)+5\right)}{e}^{3x}\\ =e^{3x}\end{array}$

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

$y=e^{-2x}\left(c_1\cos x+c_2\sin x\right)+e^{3x}$

Thus, the solution of given differential equation is$y=e^{-2x}\left(c_1\cos x+c_2\sin x\right)+e^{3x}$