Differential Operators and the Elimination Method for Systems

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 5.2 – Differential Operators and the Elimination Method for Systems

Expected Educational Results

• Objective 15–1: I can identify the linear differential operator for any polynomial function, exponential function, or sine and cosine functions.
• Objective 15–2: I can solve linear systems of ODEs using linear differential operators.

Differential Operators

Systems of Linear Equations

To convert $n^{\text{th}}$-order ODE as a system of $n$ linear differential equations:

1. Let $x_1(t)=y(t)$, $x_2(t)=y^{\,\prime}(t)$, \ldots, $x_n(t)=y^{\,(n-1)}(t)$.
2. Differentiate all $x_n(t)$. Note that $x_n^{\,\prime}(t)=y^{\,(n)}(t)=x_{n+1}$.
3. Convert all initial conditions, if any.
4. Substitute all $x_n(t)$ into the DE so that you have $n$ linear DEs in terms of $x_n^{\,\prime}(t)$.

Example 05:

Rewrite $\displaystyle y^{\,\prime\prime\prime}(t)-3y^{\,\prime\prime}(t)+y^{\,\prime}(t)-6y(t)=0$, $y(t_0)=y_0$, $y^{\,\prime}(t_0)=y_1$, $y^{\,\prime\prime}(t_0)=y_2$ into a system of linear equations.

Solution:

Let ${\color{gsured}x_1(t)=y(t)} \Rightarrow x_1^{\prime}(t)=y^{\,\prime}(t) \Rightarrow {\color{gsured}x_1^{\prime}(t)\overset{set}{=}x_2(t)}$.

Then $x_2^{\,\prime}(t)=y^{\,\prime\prime}(t)\overset{set}{=}x_3(t)\Rightarrow {\color{gsured}x_2^{\prime}(t)\overset{set}{=}x_3(t)}$

and

$x_3^{\,\prime}(t)=y^{\,\prime\prime\prime}(t)=3y^{\,\prime\prime}(t)-y^{\,\prime}(t)+6y(t)$ by solving the third-order ode for $y^{\,\prime\prime\prime}(t)$.

For the initial conditions: $y(t_0)={\color{gsured}x_1(t_0)=y_0}$, $y^{\,\prime}(t_0)={\color{gsured}x_2(t_0)=y_1}$, and $y^{\,\prime\prime}(t_0)={\color{gsured}x_3(t_0)=y_2}$

The third-order ode can be written as a system of three linear odes:

$x_1^{\prime}(t)=x_2(t), x_1(t_0)=y_0 \\ x_2^{\prime}(t)=x_3(t), x_2(t_0)=y_1 \\ x_3^{\prime}(t)=3x_3(t)-x_2(t)+6x_1(t), x_3(t_0)=y_2$

Investigation 15

Convert the following into a system of linear DEs.

1. $y^{\,\prime\prime}+2y^{\,\prime}-y=0$.
2. $y^{\,\prime\prime\prime}-2y^{\,\prime\prime}+y^{\,\prime}-3y=0$.
3. $y^{\,\prime\prime}+\cos{(t)}y^{\,\prime}+3y=t^2$, $y(0)=-2$, $y^{\,\prime}(0)=1$.

Solving Systems of Linear Equations

Example 06:

Solve the following system of linear odes:

$D[x_1] + (D+3)[x_2] = t \\ (D-1)[x_1] + (D+1)[x_2] = 0$

Solution:

Use Elimination to find $x_2(t)$.

Apply ${\color{gsured}L_1(D)=D-1}$ to the first equation; and ${\color{gsured!70!gsublue}L_2=D}$ to the second equation:

$(D-1)D[x_1] + (D-1)(D+3)[x_2] = (D-1)[t] \\ (D)(D-1)[x_1] + (D)(D+1)[x_2] = (D)[0]$

Subtract the equations. Since the linear differential operators performed on $x_1$ are the same, $x_1$ is eliminated resulting in:

$(D-1)(D+3)-(D)(D+1))[x_2]=1-t \\ \Rightarrow ((D^2+2D-3)-(D^2+D))[x_2]=1-t \\ \Rightarrow (D-3)[x_2]=1-t$

So,

$x_{2h}(t)=c_1 e^{3t}$,

$x_{2p}(t)=At+B$ [you can always find coefficients of the particular solution; here $\displaystyle A=\frac{1}{3}$, $\displaystyle B=-\frac{2}{9}$],

$\displaystyle x_{2}(t)=x_{2h}(t)+x_{2p}(t)=c_1 e^{3t}+\frac{1}{3}t-\frac{2}{9}$.

Use Elimination to find $x_1(t)$.

Apply ${\color{gsured}L_1(D)=D+1}$ to the first equation; and ${\color{gsured!70!gsublue}L_2=D+3}$ to the second equation:

$(D+1)D[x_1] + (D+1)(D+3)[x_2] = (D+1)[t] \\ (D+3)(D-1)[x_1] + (D+3)(D+1)[x_2] = (D+3)[0]$

Subtract the equations. Since the linear differential operators performed on $x_2$ are the same, $x_2$ is eliminated resulting in:

$(D+1)(D)-(D+3)(D-1))[x_1]=1+t \\ \Rightarrow ((D^2+D)-(D^2+2D-3))[x_1]=1+t \\ \Rightarrow (-D+3)[x_1]=1+t$

So,

$x_{1h}(t)=c_2 e^{3t}$ [$x_{1h}(t)$ is a different equation $x_{2h}(t)$, so no need to consider linear independence between different solutions],

$x_{1p}(t)=Ct+D$ [you can always find coefficients of the particular solution; here $\displaystyle C=\frac{1}{3}$, $\displaystyle D=\frac{4}{9}$], $\displaystyle x_{1}(t)=x_{1h}(t)+x_{1p}(t)=c_1 e^{3t}+\frac{1}{3}t+\frac{4}{9}$.

Thus, the solution to the system is:

$\displaystyle x_{1}(t)=c_1 e^{3t}+\frac{1}{3}t+\frac{4}{9}$

$\displaystyle x_{2}(t)=c_2 e^{3t}+\frac{1}{3}t-\frac{2}{9}$

Investigation 16

Solve the following systems of linear DEs.

1. $(D+1)[x_1] + D[x_2] = 2e^t \\ (D-1)[x_1] + D^2[x_2] = 0$
2. $(D + 2)[x_1] + (D + 1)[x_2] = e^{-3t} \\ (D + 4)[x_1] + (D + 3)[x_2] = e^{-2t}$

Investigation 17

Solve the following systems of linear DEs.

1. $x^{\,\prime\prime}-4x + y^{\,\prime\prime} = t^2 \\ x^{\,\prime}+ x + y^{\,\prime} = 0$
2. $x^{\,\prime\prime}+x^{\,\prime} + y^{\,\prime\prime} = 2e^t \\ x^{\,\prime\prime}-x^{\,\prime}-x +y^{\,\prime\prime}- y^{\,\prime} = 0$
3. $x^{\,\prime} = -3x + 2y, x(0) = 1 \\ y^{\,\prime} = -3x + 4y, y(0) = 1$

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Last Modified: Monday, 19 October 2020 8:23 EDT