Given that$\left\{e^x,e^{-x},e^{2x}\right\}$is a fundamental solution set for the homogeneous equation corresponding to the equation${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{,}$

determine a formula involving integrals for a particular solution.

Variation of parameters, also known as variation of constants, is a general method to solve inhomogeneous linear ordinary differential equations.

Consider the differential equation$y^{\text{'}\text{'}\text{'}}-2y^{\text{'}\text{'}}-y^{\text{'}}+2y=g\left(x\right)$

Consider the fundamental solution of the equation as,$\left\{e^x,e^{-x},e^{2x}\right\}$

Therefore the complementary solutions of the equations is$y_c=c_1{e}^x+c_2{e}^{-x}+c_{3}x{e}^{2x}$

Here we have$y_1\left(x\right)=e^x,y_2\left(x\right)=e^{-x},y_3\left(x\right)=e^{2x}$

Calculates the corresponding Wronskian is,

Apply row operation, and then Wronskian is,

$$\begin{gathered} W\left[y_1,y_2,y_3\right]=\left|\begin{array}{ccc}{e}^x& e^{-x}& e^{2x}\\ e^x& -e^x& 2e^x\\ e^x& e^{-x}& 4e^{2x}\end{array}\right| \\ W\left[y_1,y_2,y_3\right]=\left|\begin{array}{ccc}{e}^x& e^{-x}& e^{2x}\\ e^x& -e^x& 2e^x\\ 0& 0& 3e^{2x}\end{array}\right|;R_3^{\text{'}}=R_3-R_1 \\ =-6e^{2x} \end{gathered}$$

$$\begin{gathered} W_1\left(x\right)=(-1{)}^{3-1}\left|\begin{array}{cc}{e}^{-x}& e^{2x}\\ -e^{-x}& 2e^{2x}\end{array}\right| \\ =3e^x \\ {W}_2\left(x\right)=(-1{)}^{3-2}\left|\begin{array}{cc}{e}^x& e^{2x}\\ e^x& 2e^{2x}\end{array}\right| \\ =-e^{3x} \\ {W}_3\left(x\right)=(-1{)}^{3-3}\left|\begin{array}{cc}{e}^x& e^{-x}\\ e^x& -e^{-x}\end{array}\right| \\ =-2 \end{gathered}$$

We know that$v_k\left(x\right)=\int \frac{g\left(x\right)W_k\left(x\right)}{W\left[y_1,y_2,y_3\right]}dx$

Hence,

$$\begin{gathered} v_1\left(x\right)=\int \frac{g\left(x\right)3e^x}{-6e^{2x}}dx \\ =-\frac{1}{2}\int g\left(x\right)e^{-x}dx \\ {v}_2\left(x\right)=-\int \frac{g\left(x\right)e^{3x}}{-6e^{2x}}dx \\ =\frac{1}{3}\int g\left(x\right)e^{-2x}dx \\ {v}_3\left(x\right)=\int \frac{(-2)g\left(x\right)}{-6e^{2x}}dx \\ =\frac{1}{3}\int g\left(x\right)e^{-2x}dx \end{gathered}$$

Therefore the particular solution is involving integral is:

$y_p\left(x\right)=-\frac{e^x}{2}\int e^{-x}g\left(x\right)dx+\frac{e^{-x}}{6}\int e^{x}g\left(x\right)dx+\frac{e^{2x}}{3}\int e^{x}g\left(x\right)dx$