Q1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

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

Step-by-Step Solution

Answer

The particular solution is $y_p\left(x\right)=-\frac{1}{27}{e}^{2x}-\frac{2}{9}\left(\frac{3}{2}{x}^2+x\right)·e^{2x}-\frac{1}{3}{x}^2·e^{2x}$

Step 1: Definition

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

Step 2: Find complementary solution

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

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

Solving for $m$ we get value:

 $m=-1,2,2$

The  complimentary solution is $CF=c_1{e}^{-x}+c_2{e}^{2x}+c_{3}x{e}^{2x}$

Step 3: Calculate Wornkians

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

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

Find four Wronkians of determinant.

 $$\begin{gathered} W\left[y_1,y_2,y_3\right]\left(x\right)=\left|\begin{array}{c}{y}_1\\ y{\text{'}}_1\\ y{\text{'}\text{'}}_1\end{array}\begin{array}{c}{y}_2\\ y{\text{'}}_2\\ y{\text{'}\text{'}}_2\end{array}\begin{array}{c}{y}_3\\ y{\text{'}}_3\\ y{\text{'}\text{'}}_3\end{array}\right| \\ W\left[y_1,y_2,y_3\right]\left(x\right)=\left|\begin{array}{c}{}_{e^{-x}}\\ {-}_{e^{-x}}\\ {}_{e^{-x}}\end{array}\begin{array}{c}{}_{e^{2x}}\\ 2e^{2x}\\ 4e^{2x}\end{array}\begin{array}{c}x{e}^{2x}\\ 2\left(x+1\right)e^{2x}\\ 4\left(x+1\right)e^{2x}\end{array}\right| \end{gathered}$$

Take common factor out from each column.

 $W\left[y_1,y_2,y_3\right]\left(x\right)=e^{-x}·e^{2x}·e^{2x}\left|\begin{array}{c}1\\ -1\\ 1\end{array}\begin{array}{c}1\\ 2\\ 4\end{array}\begin{array}{c}x\\ \left(2x+1\right)\\ 4\left(x+1\right)\end{array}\right|$

Solving we get:

 $W_1\left(x\right)=(-1{)}^{3-1}W\left[y_2,y_3\right]\left(x\right)=e^{4x}$

$$\begin{gathered} W_2\left(x\right)=-e^x(3x+1) \\ {W}_3\left(x\right)=3e^x \end{gathered}$$

 

Step 4: Particular solution

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,

$$\begin{gathered} v_1\left(x\right)=-\int \frac{g\left(x\right)W_1\left(x\right)}{W\left[y_1,y_2,y_3\right]\left(x\right)}dx \\ {v}_1\left(x\right)=-\int \frac{e^{2x}·e^{4x}}{9e^{3x}}dx \\ {v}_1\left(x\right)=-\frac{1}{27}{e}^{3x} \\ {v}_2\left(x\right)=\int \frac{g\left(x\right)W_2\left(x\right)}{W\left[y_1,y_2,y_3\right]\left(x\right)}dx \\ {v}_2\left(x\right)=\int \frac{e^{2x}·\left\{-e^x(3x+1)\right\}}{9e^{3x}}dx \\ {v}_2\left(x\right)=-\frac{1}{9}\left(\frac{3}{2}{x}^2+x\right) \end{gathered}$$

 $$\begin{gathered} v_1\left(x\right)=-\int \frac{g\left(x\right)W_1\left(x\right)}{W\left[y_1,y_2,y_3\right]\left(x\right)}dx \\ {v}_1\left(x\right)=-\int \frac{e^{2x}·e^{4x}}{9e^{3x}}dx \\ {v}_1\left(x\right)=-\frac{1}{27}{e}^{3x} \\ {v}_2\left(x\right)=\int \frac{g\left(x\right)W_2\left(x\right)}{W\left[y_1,y_2,y_3\right]\left(x\right)}dx \\ {v}_2\left(x\right)=\int \frac{e^{2x}·\left\{-e^x(3x+1)\right\}}{9e^{3x}}dx \\ {v}_2\left(x\right)=-\frac{1}{9}\left(\frac{3}{2}{x}^2+x\right) \end{gathered}$$

$$\begin{gathered} v_3\left(x\right)=-\int \frac{g\left(x\right)W_3\left(x\right)}{W\left[y_1,y_2,y_3\right]\left(x\right)}dx \\ {v}_3\left(x\right)=-\frac{1}{3}x \end{gathered}$$

 

Thus, the particular solution is given by:

 $$\begin{gathered} y_p\left(x\right)=v_1\left(x\right)·y_1\left(x\right)+v_2\left(x\right)·y_2\left(x\right)+v_3\left(x\right)·y_3\left(x\right) \\ {y}_p\left(x\right)=-\frac{1}{27}{e}^{2x}-\frac{2}{9}\left(\frac{3}{2}{x}^2+x\right)·e^{2x}-\frac{1}{3}{x}^2·e^{2x} \end{gathered}$$