Question

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

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{3}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{3}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{x}}$

Short answer

The particular solution is$y_p=\frac{x^3{e}^x}{6}$

Step-by-step solution

Definition

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

Find complementary solution

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

The auxiliary equation is$m^3-3m^2+3m-1=0$

Solving we get$m=1,1,1$

Therefore, the complimentary solution is given by$y_c=c_1{e}^x+c_{2}x{e}^x+c_3{x}^2{e}^x$

Calculate Wornkians

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

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

$$\begin{gathered} W\left[\begin{array}{ccc}{e}^x& x{e}^x& x^2{e}^x\end{array}\right]=\left[\begin{array}{ccc}{e}^x& x{e}^x& x^2{e}^x\\ e^x& e^x\left(x+1\right)& e^x\left(x^2+2x\right)\\ e^x& e^x\left(x+2\right)& e^x\left(x^2+4x+2\right)\end{array}\right] \\ =2e^{3x} \end{gathered}$$

Calculate Wornkians

The value of wronkians$W_1,W_2,W_3$ is:

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

The particular solution is thus given by:

$$\begin{gathered} y_p=e^x\int \frac{x^2{e}^{2x}\left(e^x\right)}{2e^{3x}}dx+x{e}^x\int \frac{-2x{e}^{2x}\left(e^x\right)}{2e^{3x}}dx+x^2{e}^x\int \frac{e^{2x}·e^x}{2e^{3x}}dx \\ {y}_p=e^x\frac{x^3}{6}-\frac{x^3}{2}{e}^x+\frac{x^3}{2}{e}^x \\ =\frac{x^3{e}^x}{6} \end{gathered}$$