Reduce the system(D2 + 1)[x] − 2D[y] =

To reduce the given system of differential equations to an equivalent triangular system, we need to manipulate the equations and express the derivatives in terms of a polynomial operator $D$. The triangular system will have the form:
$P_1(D)[y]=f_1(t)$
$P_2(D)[x]+P_3(D)[y]=f_2(t)$
Let's solve the system step by step:
Step 1: Expand the differential operators
Expand the differential operators $D^2+1$ and $2D-1$ in the first equation, and $2D-1$ and $D-2$ in the second equation.
The given system is:
$(D^2+1)[x]-2D[y]=2t\text{(Equation 1)}$
$(2D-1)[x]+(D-2)[y]=7\text{(Equation 2)}$
Expanding the operators, we get:
$(D·D+1)[x]-2D[y]=2t$
$(2D-1)[x]+(D-2)[y]=7$
Step 2: Rearrange the equations
Rearrange the equations so that the derivatives are written in terms of the polynomial operator $D$.
Equation 1 becomes:
$D·(D[x])+x-2D[y]=2t$
Equation 2 becomes:
$(2D-1)[x]+(D-2)[y]=7$
Step 3: Identify the polynomial operators
By comparing the terms involving the derivatives, we can identify the polynomial operators:
$P_1(D)=D·(D[x])+1$
$P_2(D)=2D-1$
$P_3(D)=D-2$
Step 4: Express the equations in the triangular form
Rewrite the equations in the desired triangular form using the polynomial operators:
$P_1(D)[y]=f_1(t)$
$P_2(D)[x]+P_3(D)[y]=f_2(t)$
Substituting the identified operators and rearranging the terms, we have:
$(D·(D[x])+1)[y]=2t$
$(2D-1)[x]+(D-2)[y]=7$
Step 5: Simplify the expressions
Simplify the expressions further if possible.
$(D^2·x+x)[y]=2t$
$(2D-1)[x]+(D-2)[y]=7$
Step 6: Solve the system
To solve the system, we first need to solve the first equation for $x$.
$(D^2·x+x)=\frac{2t}{y}$
Differentiate both sides with respect to $t$:
$D(D^2·x+x)=D\left(\frac{2t}{y}\right)$
Simplifying the equation, we have:
$D^3·x+Dx=\frac{2}{y}$
Now we have an equation relating $x$ and its derivatives. We can substitute this equation into the second equation of the triangular system:
$(2D-1)\left(\frac{2}{y}\right)+(D-2)[y]=7$
Expanding and simplifying the equation, we have:
$\frac{4D}{y}-\frac{2}{y}+Dy-2y=7$
Combining like terms, we obtain:
$\left(\frac{4D+D^2}{y}\right)-\left(\frac{2+2y}{y}\right)=7$
To further simplify the equation, we multiply through by $y$ to eliminate the denominators:
$4D+D^2-2-2y=7y$
Rearranging the terms, we have:
$D^2+4D-9y=2+9y$
Now, the triangular system is in the form:
$P_1(D)[y]=f_1(t)$
$P_2(D)[x]+P_3(D)[y]=f_2(t)$
where:
$P_1(D)=1$
$P_2(D)=2D-1$
$P_3(D)=D-2$
$f_1(t)=\frac{2t}{y}$
$f_2(t)=2+9y$
The system is now reduced to an equivalent triangular system. To solve for $x$ and $y$, we can substitute the expression for $x$ obtained from the first equation into the second equation.
Let's recap the triangular system:
$1[y]=\frac{2t}{y}$
$(2D-1)[x]+(D-2)[y]=2+9y$
By substituting the expression for $x$ into the second equation, we have:
$(2D-1)\left(\frac{2t}{y}\right)+(D-2)[y]=2+9y$
Expanding the operators and simplifying the equation, we get:
$\frac{4Dt}{y}-\frac{2t}{y}+Dy-2y=2+9y$
Combining like terms and rearranging, we obtain:
$(\frac{4D}{y}+D)t-(\frac{2}{y}+2)=9y+2$
Now we have a simplified equation relating $t$, $y$, and their derivatives. Solving this equation will give us the solution for the system.