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7.6 Systems of Linear Differential Equations: Class Notes

7.6

Example

Use the Laplace transform to solve the given system of differential equations.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \text{(1)}\frac{dx}{dt}}& {\displaystyle =2y+e^t}\hfill \\ \hfill {\displaystyle \text{(2)}\frac{dy}{dt}}& {\displaystyle =8x-t}\hfill \end{array},x(0)=1,y(0)=1}$$

Solution

Transform both equations, substituting $x(0)=1$ and $y(0)=1$:

$${\displaystyle ℒ\{\text{(1)}\}:sX(s)-x(0)=2Y(s)+\frac{1}{s-1}}$$

$${\displaystyle ℒ\{\text{(2)}\}:sY(s)-y(0)=8X(s)-\frac{1}{s^2}}$$

Solve $ℒ\{\text{(1)}\}$ for $Y(s)$ and substitute into $ℒ\{\text{(2)}\}$.

$${\displaystyle \frac{1}{2}(sX(s)-1-\frac{1}{s-1})=Y(s)}$$

Combine the constants: ${\displaystyle -1-\frac{1}{s-1}=\frac{-s+1-1}{s-1}=-\frac{s}{s-1}}$, so

$${\displaystyle Y(s)=\frac{1}{2}(sX(s)-\frac{s}{s-1})}$$

Substitute this into $ℒ\{\text{(2)}\}$, which is ${\displaystyle sY(s)-1=8X(s)-\frac{1}{s^2}}$:

$${\displaystyle \frac{1}{2}(s^{2}X(s)-\frac{s^2}{s-1})-1=8X(s)-\frac{1}{s^2}}$$

Multiply by 2 and collect the $X(s)$ terms on the left:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle (s^2-16)X(s)}& {\displaystyle =\frac{s^2}{s-1}+2-\frac{2}{s^2}}\hfill \\ \hfill {\displaystyle (s^2-16)X(s)}& {\displaystyle =\frac{s^4+2s^2(s-1)-2(s-1)}{s^2(s-1)}}\hfill \\ \hfill {\displaystyle X(s)}& {\displaystyle =\frac{s^4+2s^3-2s^2-2s+2}{s^2(s-1)(s+4)(s-4)}}\hfill \end{array}}$$

Partial fractions:

$${\displaystyle \frac{s^4+2s^3-2s^2-2s+2}{s^2(s-1)(s+4)(s-4)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s-1}+\frac{D}{s+4}+\frac{E}{s-4}}$$

$$\begin{array}{cc}\hfill {\displaystyle s^4+2s^3-2s^2-2s+2}& {\displaystyle =As(s-1)(s^2-16)+B(s-1)(s^2-16)}\hfill \\ \hfill {\displaystyle }& {\displaystyle +C{s}^2(s^2-16)+D{s}^2(s-1)(s-4)+E{s}^2(s-1)(s+4)}\hfill \end{array}$$

Substitute the roots of the denominator, and compare the $s^4$ coefficients:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle s=0:}& {\displaystyle 2=16B\Rightarrow B=\frac{1}{8}}\hfill \\ \hfill {\displaystyle s=1:}& {\displaystyle 1=-15C\Rightarrow C=-\frac{1}{15}}\hfill \\ \hfill {\displaystyle s=4:}& {\displaystyle 346=384E\Rightarrow E=\frac{173}{192}}\hfill \\ \hfill {\displaystyle s=-4:}& {\displaystyle 106=640D\Rightarrow D=\frac{53}{320}}\hfill \\ \hfill {\displaystyle s^4:}& {\displaystyle 1=A+C+D+E\Rightarrow A=1+\frac{1}{15}-\frac{173}{192}-\frac{53}{320}=0}\hfill \end{array}}$$

$${\displaystyle X(s)=\frac{1/8}{s^2}-\frac{1/15}{s-1}+\frac{53/320}{s+4}+\frac{173/192}{s-4}}$$

Rather than transforming back again, solve $\text{(1)}$ for $y$ and use $x(t)$:

$${\displaystyle y(t)=\frac{1}{2}x\prime (t)-\frac{1}{2}{e}^t}$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \text{(1)}\frac{d^{2}x}{d{t}^2}+\frac{dx}{dt}+\frac{dy}{dt}}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle \text{(2)}\frac{d^{2}y}{d{t}^2}+\frac{dy}{dt}-4\frac{dx}{dt}}& {\displaystyle =0}\hfill \end{array}}$$

$$x(0)=1,x\prime (0)=0,y(0)=-1,y\prime (0)=5$$

Solution

Transform both equations. The terms that come from $x(0)=1$, $y(0)=-1$, and $y\prime (0)=5$ are highlighted to match: $x(0)=1$ gives $-s$ and $-1$ in the first equation and $+4$ in the second; $y(0)=-1$ gives $+1$ in the first and $+s$ and $+1$ in the second; $y\prime (0)=5$ gives $-5$ in the second.

$$ℒ\{\text{(1)}\}:s^{2}X(s)-s+sX(s)-1+sY(s)+1=0$$

$$ℒ\{\text{(2)}\}:s^{2}Y(s)+s-5+sY(s)+1-4sX(s)+4=0$$

In $ℒ\{\text{(1)}\}$ the constants $-1+1$ cancel, and in $ℒ\{\text{(2)}\}$ the constants $-5+1+4$ cancel. Then divide each equation by $s$:

$$(s^2+s)X(s)+sY(s)=s\Rightarrow (s+1)X(s)+Y(s)=1$$

$$(s^2+s)Y(s)-4sX(s)=-s\Rightarrow (s+1)Y(s)-4X(s)=-1$$

Elimination: eliminate $Y(s)$. Multiply the first equation by $s+1$, rewrite the second as $4X(s)-(s+1)Y(s)=1$, and add, so the $Y(s)$ terms cancel:

$$\begin{array}{cc}\hfill {\displaystyle (s+1{)}^{2}X(s)+(s+1)Y(s)}& {\displaystyle =s+1}\hfill \\ \hfill {\displaystyle 4X(s)-(s+1)Y(s)}& {\displaystyle =1}\hfill \end{array}$$

Adding:

$$[(s+1{)}^2+4]X(s)=s+2$$

$${\displaystyle X(s)=\frac{s+2}{(s+1{)}^2+4}}$$

Write the numerator in terms of $s+1$ to match the denominator, and give the second piece the constant $k=2$ it needs for a sine transform:

$${\displaystyle X(s)=\frac{s+1}{(s+1{)}^2+4}+\frac{1}{2}·\frac{1·2}{(s+1{)}^2+4}}$$

$${\displaystyle x(t)=e^{-t}\cos 2t+\frac{1}{2}{e}^{-t}\sin 2t}$$

The first transformed equation gives $Y(s)=1-(s+1)X(s)$. Substitute $X(s)$:

$${\displaystyle Y(s)=1-\frac{(s+1)(s+2)}{(s+1{)}^2+4}}$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle Y(s)}& {\displaystyle =\frac{s^2+2s+5-s^2-3s-2}{(s+1{)}^2+4}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{-s+3}{(s+1{)}^2+4}}\hfill \end{array}}$$

Again write the numerator in terms of $s+1$: $-s+3=-(s+1)+4$, and $4=2·2$ supplies $k=2$:

$${\displaystyle Y(s)=-\frac{s+1}{(s+1{)}^2+4}+\frac{2·2}{(s+1{)}^2+4}}$$