Question

Examine and discuss the following system:

$$\begin{gathered} \mathit{D}\mathit{x}\mathbf{-}\mathbf{2}\mathit{D}\mathit{y}\mathbf{=}{\mathit{t}}^{\mathbf{2}} \\ \mathbf{(}\mathit{D}\mathbf{+}\mathbf{1}\mathbf{)}\mathit{x}\mathbf{-}\mathbf{2}\mathbf{(}\mathit{D}\mathbf{+}\mathbf{1}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{1}\mathbf{.} \end{gathered}$$

Short answer

The system of equations has No solution.

Step-by-step solution

To Find the differential equation

The standard system of equations is given as

$$\begin{gathered} a_{1}x+b_{1}y=c_1 \\ {a}_{2}x+b_{2}y=c_2 \end{gathered}$$

Comparing the given system of equations with the standard, we get

$$\begin{gathered} a_1=D,a_2=(D+1) \\ {b}_1=-2D,b_2=-2(D+1) \\ {c}_1=t^2,c_2=1 \end{gathered}$$

Using the above facts, we have that

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Final Answer

Therefore, the given system of equations has NO solution.