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4.9 Solving Systems of Linear DEs by Elimination: Class Notes

4.9a

Differential operator: for $y=y(x)$,

$${\displaystyle Dy=\frac{dy}{dx}\text{and}{D}^{2}y=\frac{d^{2}y}{d{x}^2}.}$$

$y\prime +5y$ can be written as $Dy+5y=(D+5)y$. We can operate on this using $D$:

$$D(D+5)y=D(Dy+5y)=D^{2}y+5Dy=(D^2+5D)y$$

Further, we can operate on $(D+5)y$ using $D+2$:

$$\begin{array}{cc}\hfill {\displaystyle (D+2)(D+5)y}& {\displaystyle =D(D+5)y+2(D+5)y}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(D^2+5D)y+(2D+10)y}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(D^2+7D+10)y}\hfill \end{array}$$

For example, $y\prime +5y=e^{3x}$ becomes $Dy+5y=e^{3x}$. Applying $D+2$ to both sides:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle (D+2)(D+5)y}& {\displaystyle =(D+2)e^{3x}=\frac{d}{dx}\left(e^{3x}\right)+2e^{3x}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =3e^{3x}+2e^{3x}=5e^{3x}}\hfill \end{array}}$$

4.9b & 4.9c

The DE $y\prime +2y+3x=-1$ can be written in terms of the differential operator $D$, where $D$ represents a derivative. In this form, the DE becomes $Dy+2y+3x=-1$. This can also be written $(D+2)y+3x=-1$. Our work in this lesson makes use of this notation. For this lesson, $x$ and $y$ are functions of $t$, so ${\displaystyle Dx=\frac{dx}{dt}}$, ${\displaystyle Dy=\frac{dy}{dt}}$, and ${\displaystyle D^{2}x=\frac{d^{2}x}{d{t}^2}}$, etc.

Solve the given system of differential equations by elimination.

Example

$$\begin{array}{cc}\hfill {\displaystyle (1)(D+1)x+(D-1)y}& {\displaystyle =2}\hfill \\ \hfill {\displaystyle (2)3x+(D+2)y}& {\displaystyle =-1}\hfill \end{array}$$

Solution

We can eliminate $x$ or $y$ by applying appropriate differential operators and subtracting. Eliminate $x$ by multiplying (1) by 3, applying $D+1$ to (2), and subtracting:

$$\begin{array}{cc}\hfill {\displaystyle 3(D+1)x+3(D-1)y}& {\displaystyle =6}\hfill \\ \hfill {\displaystyle 3(D+1)x+(D+1)(D+2)y}& {\displaystyle =(D+1)(-1)=-1}\hfill \\ \hfill {\displaystyle [3(D-1)-(D^2+3D+2)]y}& {\displaystyle =6+1}\hfill \\ \hfill {\displaystyle (-D^2-5)y}& {\displaystyle =7}\hfill \\ \hfill {\displaystyle (D^2+5)y}& {\displaystyle =-7}\hfill \end{array}$$

The derivative of a constant is 0, so $(D+1)(-1)=0-1=-1$.

That is, $y″+5y=-7$. The auxiliary equation $m^2+5=0$ gives $m=\pm \sqrt{5}i$, so

$$y_c(t)=c_1\cos \sqrt{5}t+c_2\sin \sqrt{5}t.$$

Since $g(t)=-7$ is a constant, assume $y_p=A$. Then $0+5A=-7$, so ${\displaystyle A=-\frac{7}{5}}$:

$${\displaystyle y(t)=c_1\cos \sqrt{5}t+c_2\sin \sqrt{5}t-\frac{7}{5}}$$

To find $x$, solve (2) for it:

$${\displaystyle 3x+(D+2)y=-1⟹x=-\frac{1}{3}-\frac{1}{3}(D+2)y=-\frac{1}{3}-\frac{1}{3}Dy-\frac{2}{3}y}$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle -\frac{1}{3}Dy}& {\displaystyle =\frac{\sqrt{5}}{3}{c}_1\sin \sqrt{5}t-\frac{\sqrt{5}}{3}{c}_2\cos \sqrt{5}t}\hfill \\ \hfill {\displaystyle -\frac{2}{3}y}& {\displaystyle =-\frac{2}{3}{c}_1\cos \sqrt{5}t-\frac{2}{3}{c}_2\sin \sqrt{5}t+\frac{14}{15}}\hfill \end{array}}$$

Add these to ${\displaystyle -\frac{1}{3}}$ and collect the cosine and sine terms (the constants give ${\displaystyle -\frac{1}{3}+\frac{14}{15}=\frac{3}{5}}$):

Alternative approach for finding $x$ (assuming we already found $y$): now eliminate $y$ by applying $D+2$ to (1) and $D-1$ to (2), then subtracting:

$$\begin{array}{cc}\hfill {\displaystyle (D+2)(D+1)x+(D+2)(D-1)y}& {\displaystyle =(D+2)2=4}\hfill \\ \hfill {\displaystyle 3(D-1)x+(D-1)(D+2)y}& {\displaystyle =(D-1)(-1)=1}\hfill \\ \hfill {\displaystyle (D^2+3D+2-3D+3)x}& {\displaystyle =4-1}\hfill \\ \hfill {\displaystyle (D^2+5)x}& {\displaystyle =3}\hfill \end{array}$$

So $m=\pm \sqrt{5}i$ and $x_c(t)=c_3\cos \sqrt{5}t+c_4\sin \sqrt{5}t$. With $x_p=A$, $5A=3$, so ${\displaystyle A=\frac{3}{5}}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle x(t)}& {\displaystyle =c_3\cos \sqrt{5}t+c_4\sin \sqrt{5}t+\frac{3}{5}}\hfill \\ \hfill {\displaystyle y(t)}& {\displaystyle =c_1\cos \sqrt{5}t+c_2\sin \sqrt{5}t-\frac{7}{5}}\hfill \end{array}}$$

We have too many constants: $c_1$, $c_2$, $c_3$, and $c_4$ are constrained by the system, so there are relationships among them. To reconcile them, substitute $x(t)$ and $y(t)$ into (2) (an arbitrary choice; (1) works too):

$${\displaystyle \begin{array}{c}{\displaystyle 3(c_3\cos \sqrt{5}t+c_4\sin \sqrt{5}t+\frac{3}{5})+(D+2)(c_1\cos \sqrt{5}t+c_2\sin \sqrt{5}t-\frac{7}{5})=-1}\hfill \\ {\displaystyle 3c_3\cos \sqrt{5}t+3c_4\sin \sqrt{5}t+\frac{9}{5}-\sqrt{5}{c}_1\sin \sqrt{5}t+\sqrt{5}{c}_2\cos \sqrt{5}t}\hfill \\ {\displaystyle +2c_1\cos \sqrt{5}t+2c_2\sin \sqrt{5}t-\frac{14}{5}=-1}\hfill \end{array}}$$

Match the cosine terms, the sine terms, and the constants on the two sides:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle 3c_3+\sqrt{5}{c}_2+2c_1}& {\displaystyle =0⟹c_3=-\frac{2}{3}{c}_1-\frac{\sqrt{5}}{3}{c}_2}\hfill \\ \hfill {\displaystyle 3c_4-\sqrt{5}{c}_1+2c_2}& {\displaystyle =0⟹c_4=\frac{\sqrt{5}}{3}{c}_1-\frac{2}{3}{c}_2}\hfill \\ \hfill {\displaystyle \frac{9}{5}-\frac{14}{5}}& {\displaystyle =-1}\hfill \end{array}}$$

Substituting these for $c_3$ and $c_4$ gives the same $x(t)$ as before.

4.9d

Example

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{dx}{dt}+\frac{dy}{dt}}& {\displaystyle =e^t}\hfill \\ \hfill {\displaystyle -\frac{d^{2}x}{d{t}^2}+\frac{dx}{dt}+x+y}& {\displaystyle =0}\hfill \end{array}}$$

Solution

In operator notation:

$$\begin{array}{cc}\hfill {\displaystyle (1)Dx+Dy}& {\displaystyle =e^t}\hfill \\ \hfill {\displaystyle (2)(-D^2+D+1)x+y}& {\displaystyle =0}\hfill \end{array}$$

Eliminate $y$ with (1) − $D$(2):

$$\begin{array}{cc}\hfill {\displaystyle Dx-D(-D^2+D+1)x}& {\displaystyle =e^t}\hfill \\ \hfill {\displaystyle Dx+D^{3}x-D^{2}x-Dx}& {\displaystyle =e^t}\hfill \\ \hfill {\displaystyle (D^3-D^2)x}& {\displaystyle =e^t}\hfill \end{array}$$

The auxiliary equation is $m^3-m^2=0$, or $m^2(m-1)=0$, so $m=0$ (repeated) and $m=1$:

$$x_c(t)=c_1+c_{2}t+c_3{e}^t$$

Assume $x_p=At{e}^t$ (avoiding redundancy, since $e^t$ is already in $x_c$). Then

$$x_p\prime =(A+At)e^t,x_p″=(2A+At)e^t,x_p‴=(3A+At)e^t.$$

Substituting into $x‴-x″=e^t$ and omitting the common factor $e^t$:

$$3A+At-(2A+At)=1⟹A=1$$

$$x_p=t{e}^t⟹x(t)=c_1+c_{2}t+c_3{e}^t+t{e}^t$$

To find $y$, solve (2) for it: $-D^{2}x+Dx+x+y=0$, so $y=D^{2}x-Dx-x$. With

$$Dx=c_2+c_3{e}^t+e^t+t{e}^t,D^{2}x=c_3{e}^t+2e^t+t{e}^t,$$

$$\begin{array}{cc}\hfill {\displaystyle y}& {\displaystyle =(c_3{e}^t+2e^t+t{e}^t)-(c_2+c_3{e}^t+e^t+t{e}^t)}\hfill \\ \hfill {\displaystyle }& {\displaystyle -(c_1+c_{2}t+c_3{e}^t+t{e}^t)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =-c_1-c_2-c_{2}t-c_3{e}^t-t{e}^t+e^t}\hfill \end{array}$$

Example

$$\begin{array}{cc}\hfill {\displaystyle (1)Dx+z}& {\displaystyle =e^t}\hfill \\ \hfill {\displaystyle (2)(D-1)x+Dy+Dz}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (3)x+2y+Dz}& {\displaystyle =e^t}\hfill \end{array}$$

Solution

Eliminate $y$ with 2(2) − $D$(3):

$$\begin{array}{cc}\hfill {\displaystyle (2D-2)x+2Dy+2Dz}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle Dx+2Dy+D^{2}z}& {\displaystyle =D{e}^t=e^t}\hfill \\ \hfill {\displaystyle (4)(D-2)x+(-D^2+2D)z}& {\displaystyle =-e^t}\hfill \end{array}$$

Eliminate $z$ with (4) + $(D^2-2D)$(1):

$$\begin{array}{cc}\hfill {\displaystyle (D-2)x+(-D^2+2D)z}& {\displaystyle =-e^t}\hfill \\ \hfill {\displaystyle (D^3-2D^2)x+(D^2-2D)z}& {\displaystyle =(D^2-2D)e^t=e^t-2e^t=-e^t}\hfill \\ \hfill {\displaystyle (D^3-2D^2+D-2)x}& {\displaystyle =-2e^t}\hfill \end{array}$$

The auxiliary equation factors by grouping: $m^2(m-2)+(m-2)=(m^2+1)(m-2)=0$, so $m=2,\pm i$:

$$x_c=c_1{e}^{2t}+c_2\cos t+c_3\sin t$$

With $x_p=A{e}^t$, every derivative of $x_p$ is $A{e}^t$, so $A-2A+A-2A=-2$ and $A=1$:

$$x(t)=c_1{e}^{2t}+c_2\cos t+c_3\sin t+e^t$$

From (1), with $Dx=2c_1{e}^{2t}-c_2\sin t+c_3\cos t+e^t$:

$$z=e^t-Dx=e^t-(2c_1{e}^{2t}-c_2\sin t+c_3\cos t+e^t)=-2c_1{e}^{2t}+c_2\sin t-c_3\cos t$$

From (3), ${\displaystyle y=\frac{1}{2}{e}^t-\frac{1}{2}x-\frac{1}{2}Dz}$, with $Dz=-4c_1{e}^{2t}+c_2\cos t+c_3\sin t$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle y}& {\displaystyle =\frac{1}{2}{e}^t-\frac{1}{2}{c}_1{e}^{2t}-\frac{1}{2}{c}_2\cos t-\frac{1}{2}{c}_3\sin t-\frac{1}{2}{e}^t}\hfill \\ \hfill {\displaystyle }& {\displaystyle +2c_1{e}^{2t}-\frac{1}{2}{c}_2\cos t-\frac{1}{2}{c}_3\sin t}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{3}{2}{c}_1{e}^{2t}-c_2\cos t-c_3\sin t}\hfill \end{array}}$$

4.9e

Example

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{dx}{dt}}& {\displaystyle =-x+z}\hfill \\ \hfill {\displaystyle \frac{dy}{dt}}& {\displaystyle =-y+z}\hfill \\ \hfill {\displaystyle \frac{dz}{dt}}& {\displaystyle =-x+y}\hfill \end{array}}$$

Solution

In operator notation:

$$\begin{array}{cc}\hfill {\displaystyle (1)(D+1)x-z}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (2)(D+1)y-z}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (3)x-y+Dz}& {\displaystyle =0}\hfill \end{array}$$

Eliminate $y$ with (2) + $(D+1)$(3):

$$\begin{array}{cc}\hfill {\displaystyle (D+1)y-z}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (D+1)x-(D+1)y+(D^2+D)z}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (4)(D+1)x+(D^2+D-1)z}& {\displaystyle =0}\hfill \end{array}$$

Then (4) − (1) eliminates $x$:

$$(D^2+D)z=0⟹m(m+1)=0⟹m=0,-1$$

$$z(t)=c_1+c_2{e}^{-t}$$

Eliminate $z$ from (1) and (4) with $(D^2+D-1)$(1) + (4):

$$\begin{array}{cc}\hfill {\displaystyle (D^2+D-1)(D+1)x-(D^2+D-1)z}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (D+1)x+(D^2+D-1)z}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle [(D^2+D-1)(D+1)+(D+1)]x}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (D+1)(D^2+D-1+1)x}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle D(D+1{)}^{2}x}& {\displaystyle =0}\hfill \end{array}$$

So $m=0$ and $m=-1$ (repeated):

$$x(t)=c_3+c_4{e}^{-t}+c_{5}t{e}^{-t}$$

From (3), $y=x+Dz$:

$$y=c_3+c_4{e}^{-t}+c_{5}t{e}^{-t}-c_2{e}^{-t}=c_3+(c_4-c_2)e^{-t}+c_{5}t{e}^{-t}$$

To reconcile the constants, use (1): $(D+1)x=z$, that is, $Dx+x=z$:

$$\begin{array}{c}{\displaystyle \underset{Dx}{\underbrace{-c_4{e}^{-t}+c_5(1-t)e^{-t}}}+\underset{x}{\underbrace{c_3+c_4{e}^{-t}+c_{5}t{e}^{-t}}}=\underset{z}{\underbrace{c_1+c_2{e}^{-t}}}}\hfill \end{array}$$

Comparing the $e^{-t}$ terms: $-c_4+c_5+c_4=c_2$, so $c_2=c_5$. Comparing constants: $c_3=c_1$.