In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.

${\mathit{z}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{3}{\mathit{z}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{4}\mathit{z}\mathbf{=}{\mathit{e}}^{\mathbf{2}\mathbf{x}}$

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

The given equation is:$z^{\text{'}\text{'}\text{'}}+3z^{\text{'}\text{'}}-4z=e^{2x}$

The auxiliary equation is$m^3+3m^2-4=0$

Simplifying we get

$$\begin{gathered} m^3+3m^2-4=0 \\ (m-1)(m+2{)}^2=0 \\ m=1,-2,-2 \end{gathered}$$

Therefore, the complimentary solution is given by$Z_c=c_1{e}^x+c_2{e}^{-2x}+c_{3}x{e}^{-2x}$

Compare with.$CF=c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)+c_3{y}_3\left(x\right)$

$y_1\left(x\right)=e^x,y_2\left(x\right)=e^{-2x}\text{and}{y}_3\left(x\right)=x{e}^{-2x}$

$$\begin{gathered} W\left(x\right)=\left|\begin{array}{ccc}{e}^x& e^{-2x}& x{e}^{-2x}\\ {\left(e^x\right)}^0& {\left(e^{2x}\right)}^0& \left(x{e}^{-2x}\right)\\ {\left(e^x\right)}^{*}& \left(e^{-2x}\right)*& \left(x{e}^{-2x}\right)*\end{array}\right| \\ =e^x\left(e^{-4x}(-8x+8+8x-4)\right)-e^{-2x}\left(e^{-x}(4x-4+2x-1)\right)+x{e}^{-2x}\left(4e^{-x}+2e^{-x}\right) \\ =4e^{-3x}-6x{e}^{-3x}+5e^{-3x}+6x{e}^{-3x} \\ =9e^{-3x} \end{gathered}$$

$$\begin{gathered} W_1=(-1{)}^{3-1}W\left(e^{-2x},x{e}^{-2x}\right) \\ =\left|\begin{array}{cc}{e}^{-2x}& x{e}^{-2x}\\ -2e^{-2x}& e^{-2x}\left(1-2x\right)\end{array}\right| \\ =e^{-4x} \\ {W}_2=(-1{)}^{3-2}W\left(e^x,x{e}^{-2x}\right) \\ =-\left|\begin{array}{cc}{e}^x& x{e}^{-2x}\\ e^x& e^{-2x}\left(1-2x\right)\end{array}\right| \\ =-e^{-x}(1-3x) \end{gathered}$$

And the value$w_3$is,

$$\begin{gathered} W_3=(-1{)}^{3-3}W\left(e^x,e^{-2x}\right) \\ =\left|\begin{array}{cc}{e}^x& e^{-2x}\\ e^x& -2e^{-2x}\end{array}\right| \\ =-3e^{-x} \end{gathered}$$

The particular solution is given by$y_p\left(x\right)=y_1\left(x\right)·v_1\left(x\right)+y_2\left(x\right)·v_2\left(x\right)+y_3\left(x\right)·v_3\left(x\right)$

Here,

$$\begin{gathered} V_1=\int \frac{W_1\left(x\right)f\left(x\right)}{W\left(x\right)}dx \\ =\frac{e^x}{9} \\ {V}_2=\int \frac{W_2\left(x\right)f\left(x\right)}{W\left(x\right)}dx \\ =\int \frac{-e^{-x}(1-3x)e^{2x}}{9e^{-3x}}dx \\ =-\frac{7e^{4x}}{144}+\frac{x{e}^{4x}}{12} \end{gathered}$$

And

$$\begin{gathered} V_3=\int \frac{W_3\left(x\right)f\left(x\right)}{W\left(x\right)}dx \\ =-\frac{e^{4x}}{12} \end{gathered}$$

Therefore, the particular integral is given by,

$$\begin{gathered} Z_p=V_1{e}^x+V_2{e}^{-2x}+V_{3}x{e}^{-2x} \\ =e^x\frac{e^x}{9}+\left(\frac{e^{4x}}{144}+\frac{x{e}^{4x}}{12}\right)e^{-2x}+\left(-\frac{e^{4x}}{12}\right)x{e}^{-2x} \\ =\frac{e^{2x}}{16} \end{gathered}$$

Therefore the particular solution is$Z_p=\frac{1}{16}{e}^{2x}$