91Ó°ÊÓ

The differential equation is$y^{\text{'}\text{'}}−4y^{\text{'}}+4y=6e^{2x}$.

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$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{a}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{b}\mathbf{x}}\mathbf{.}$

$\begin{array}{c}{y}_p=\frac{6e^{2x}}{\left(D^2−5D+6\right)}\\ =\frac{6e^{2x}}{{(D+2)}^2−4(D+2)+4}\\ =\frac{6e^{2x}}{D^2}\\ =\frac{6e^{2x}\left(x^2\right)}{2}\end{array}$Consider the equation.

$y^{\text{'}\text{'}}−4y^{\text{'}}+4y=6e^{2x}$

The above equation is a non-homogenous equation.

Substitute the values in above equation.

$\left(D^2−4D+4\right)y=6e^{2x}$

The auxiliary equation is,

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

The solution for m = 2, 2 is,

$y_c=e^{2x}\left(c_{1}x+c_2\right)$

The solution of equation is,

$y=y_c+y_p……………\mathrm{..}\left(1\right)$

The value of y_p is,

$\begin{array}{c}{y}_p=\frac{6e^{2x}}{\left(D^2−5D+6\right)}\\ =\frac{6e^{2x}}{{(D+2)}^2−4(D+2)+4}\\ =\frac{6e^{2x}}{D^2}\\ =\frac{6e^{2x}\left(x^2\right)}{2}\end{array}$

Solve the equation further

$y_p=3x^2{e}^{2x}$

Substitute the values of y_pand y_c in Equation (1).

$y=e^{2x}\left(c_{1}x+c_2\right)+3x^2{e}^{2x}$

Thus, the solution of given differential equation is $y=e^{2x}\left(c_{1}x+c_2\right)+3x^2{e}^{2x}$.