Question

Use the operator method described in this section to find the general solution of each of the linear systems in Exercises 1-26. $$ \begin{aligned} &x^{\prime}+y^{\prime}-x-3 y=3 t \\ &x^{\prime}+2 y^{\prime}-2 x-3 y=1 \end{aligned} $$

Short answer

The general solution for the given linear system of ODEs is: $$ \begin{aligned} x &= t - (e^{t}-1) \\ y &= e^{3t}(e^{-2t} - 1) \end{aligned} $$

Step-by-step solution

Express the given equations in the operator form

We can write the given equations in the operator form as: $$ \begin{aligned} &(D + 1)x + (D - 3)y = 3t \\ &(D - 2)x + (D - 3)y = 1 \end{aligned} $$ where \(D\) is the differentiation operator.

Eliminate one of the variables

To eliminate one of the variables, we multiply the first equation by 2 and then subtract the second equation, which gives us: $$ [(2D - D)(D + 1)]x =(6t - 2) $$ Simplifying this equation, we get: $$ (D^2 -D)x = 6t - 2 $$

Solve for x using inverse operator method

Now we solve the previous equation for x using the inverse operator method. Applying the inverse operator method, we get: $$ x = \frac{1}{D^2-D}\big(6t-2\big) $$ Now we find the inverse operator \(C = \frac{1}{D (D-1)}\), using partial fraction decomposition, we have: $$ C = \frac{A}{D} + \frac{B}{D-1} $$ To find \(A\) and \(B\), we multiply both sides by \(D(D-1)\), then we obtain: $$ 1 = A(D-1) + B(D) $$ To find the values of \(A\) and \(B\), we equate each coefficient: A = -B At D=0, -B = 1 => B = -1, and A = 1 So, the operator C can be rewritten as: $$ C = \frac{1}{D} - \frac{1}{D-1} $$ Applying operator C to \(6t-2\), we get the general solution for x: $$ x = t - (e^{t}-1) $$

Solve for y using back-substitution

We can find y from the first equation since we already have x: $$ (D - 3)y = 3t - (D + 1)x $$ Substituting the expression for x we found previously: $$ (D - 3)y = 3t - (D + 1)(t - (e^{t}-1)) $$ $$ (D - 3)y = 3t - (Dt-D+ t - e^{t}+1) $$ $$ (D - 3)y = 2e^{t} - 1 $$ Now we apply the inverse operator method again to find the general solution for y: $$ y = \frac{1}{D - 3}\left(2e^{t}-1\right) $$ Applying the inverse operator \(3 - D\), we get the general solution for y: $$ y = e^{3t}(e^{-2t} - 1) $$

Combine the solutions

Finally, combining the solutions for x and y, we get the general solution for the given linear system of ODEs: $$ \begin{aligned} x &= t - (e^{t}-1) \\ y &= e^{3t}(e^{-2t} - 1) \end{aligned} $$

Key concepts

Operator Method

The operator method is a technique used to solve linear systems of ordinary differential equations (ODEs). This approach transforms an ODE into a form that is often easier to manipulate and solve. Imagine the differentiation process as an 'operation' that can be applied to a function, much like multiplication or division. In this context, we denote the differentiation operator as \(D\). For example, if \(y\) is a function of time, \(t\), then \(Dy\) represents the derivative \(y'\).

Using the operator method, we turn the derivatives in an ODE into expressions involving the operator \(D\). This abstract representation allows for algebraic manipulation, making it possible to apply techniques such as elimination, substitution, or even inverse operations – which can simplify solving the system significantly. In the given exercise, the operator method was employed to convert a system of differential equations into operator form, setting the stage for a streamlined solution process, as you'll see in the following steps.

General Solution of ODEs

The general solution of an ODE incorporates all possible solutions to the equation. It generally includes a number of arbitrary constants equal to the order of the differential equation. The goal of solving an ODE is to find a function or a set of functions that satisfy the equation for all points in a particular interval. In a linear system of ODEs, like the one presented in the exercise, the general solution often involves finding solutions for each variable that satisfy all the equations in the system simultaneously.

A critical aspect of finding the general solution is handling the homogeneous and particular parts of the equations separately. The homogeneous solution corresponds to the solution of the equation when set equal to zero, while the particular solution is specific to the non-homogeneous part – which, in our case, includes the terms involving \(t\). By combining the homogeneous and particular solutions appropriately, we address the full complexity of the ODE and arrive at the most comprehensive answer possible.

Inverse Operator Method

The inverse operator method is a sophisticated mathematical technique that is instrumental in unraveling differential equations. Comparable to inverting a number in basic algebra, the inverse operator is used to 'undo' a differential operation. This powerful method allows for finding a particular solution to a non-homogeneous differential equation.

To apply the inverse operator, we first express the equation with operator notation and then determine the explicit form of the inverse of the operator. As part of the process, partial fraction decomposition might come into play to break down complex operator expressions into simpler, more manageable parts. In the given exercise solution, this methodology enabled us to transform the equation \((D^2 - D)x = 6t - 2\) into a form that allowed for direct computation of the function \(x\). The subsequent use of the inverse operator produced the actual function that constitutes the general solution for \(x\).

Partial Fraction Decomposition

Partial fraction decomposition is an essential algebraic tool, particularly valuable when working with rational expressions involving polynomials. This technique breaks down complex fractions into a sum of simpler fractions, where each denominator is a factor of the original denominator. This process greatly simplifies integration or inverse operations involving complex rational expressions.

In the context of solving differential equations using inverse operators, partial fraction decomposition is used to express the inverse operator as a sum of simpler operators. This is done so that we can then directly apply these simpler operators to the functions or constants in our equation. During our exercise, the decomposition process transformed the inverse operator \(\frac{1}{D (D-1)}\) into \(\frac{1}{D} - \frac{1}{D-1}\) with coefficients \(A\) and \(B\) that were determined by solving the equation \(1 = A(D-1) + B(D)\). This decomposition was critical to finding a solution that could be more easily processed to yield the general solution for the variable \(x\).