Q2E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Show that the operator (D-1)(D+2) is the same as the operator ${\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}$.

Step-by-Step Solution

Answer

Thus, it is proved that the operator $\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{2}\right)$ is the same as the operator ${\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}$

Step 1: General form

Elimination Procedure for 2 × 2 Systems:

 

To find a general solution for the system

 $$\begin{gathered} {\mathbf{L}}_{\mathbf{1}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{1}}\mathbf{,} \\ {\mathbf{L}}_{\mathbf{3}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{4}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{2}}\mathbf{,} \end{gathered}$$

a. Make sure that the system is written in operator form.

 

b. Eliminate one of the variables, say, y, and solve the resulting equation for x(t). If the system is degenerating, stop! A separate analysis is required to determine whether or not there are solutions.

 

c. (Shortcut) If possible, use the system to derive an equation that involves y(t) but not its derivatives. [Otherwise, go to step (d).] Substitute the found expression for x(t) into this equation to get a formula for y(t). The expressions for x(t), y(t) gives the desired general solution.

 

d. Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants----in fact, twice as many as needed.]

 

e. Remove the extra constants by substituting the expressions for x(t) and y(t) into one or both of the equations in the system. Write the expressions for x(t) and y(t) in terms of the remaining constants.

Step 2: Evaluate the given equation

Given that, $\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{2}\right)$ where $\mathbf{D}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}$.

 

To prove give equation and ${\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}$ are same.

 

Let us take, $\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{2}\right)\left[\mathbf{y}\right]$ it and solve it.

 

Then,

 $$\begin{gathered} \left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{2}\right)\left[\mathbf{y}\right]\mathbf{=}\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\right] \\ \mathbf{=}\mathbf{D}\left[\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\right]\mathbf{-}\mathbf{1}\left[\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\right] \\ \mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y} \\ \mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y} \\ \mathbf{=}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}\right)\left[\mathbf{y}\right] \end{gathered}$$

 Hence, it is proved that the operator $\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{2}\right)$ is the same as the operator ${\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}$.