Kansas State University

Discussion

Paradigm

The only trick here is that both equations need to be transformed. $$ \eqalign { s{\mathcal L}\{x\}-1&=3{\mathcal L}\{x\}+5{\mathcal L}\{y\} \cr s{\mathcal L}\{y\}-2&=3{\mathcal L}\{x\}+{\mathcal L}\{y\} \cr} $$ Step 2: Solve for ${\mathcal L}\{x\}$ and ${\mathcal L}\{y\}$.

Here we have to solve two equations in two unknowns instead of just one equation in one unknown, so the algebra is messier. First we collect the unknowns ${\mathcal L}\{x\}$ and ${\mathcal L}\{y\}$ on the same side of the equation. $$ \eqalign { (s-3){\mathcal L}\{x\}-5{\mathcal L}\{y\}&=1 \cr -3{\mathcal L}\{x\}+(s-1){\mathcal L}\{y\}&=2 \cr} $$ Next we multiply the top equation by $3$ and the bottom equation by $s-3$ so the two ${\mathcal L}\{x\}$ terms will have the same coefficients. $$ \eqalign { 3(s-3){\mathcal L}\{x\}-15{\mathcal L}\{y\}&=3 \cr -3(s-3){\mathcal L}\{x\}+(s-3)(s-1){\mathcal L}\{y\}&=2(s-3) \cr } $$ Now we add the equations to get the ${\mathcal L}\{x\}$ term to drop out. $$[(s-3)(s-1)-15]{\mathcal L}\{y\}=2(s-3)+3$$ Then we solve for ${\mathcal L}\{y\}$. $${\mathcal L}\{y\}=\frac{2s-3}{s^2-4s-12}$$ Now we plug our formula for ${\mathcal L}\{y\}$ back into our first equation. $$(s-3){\mathcal L}\{x\}-5\frac{2s-3}{s^2-4s-12}=1$$ Then we solve for ${\mathcal L}\{x\}$. $$ \eqalign { (s-3){\mathcal L}\{x\}&=1+5\frac{2s-3}{s^2-4s-12} \cr {\mathcal L}\{x\}&=\frac{1}{s-3}+5\frac{2s-3}{(s-3)(s^2-4s-12)} \cr {\mathcal L}\{x\}&=\frac{(s^2-4s-12)+(10s-15)}{(s-3)(s^2-4s-12)} \cr {\mathcal L}\{x\}&=\frac{s^2+6s-27}{(s-3)(s^2-4s-12)} \cr {\mathcal L}\{x\}&=\frac{(s+9)(s-3)}{(s-3)(s-6)(s+2)} \cr {\mathcal L}\{x\}&=\frac{s+9}{(s-6)(s+2)} } $$ Step 3: Take inverse Laplace transforms to find $x(t)$ and $y(t)$.

At this stage we have separate formulas for ${\mathcal L}\{x\}$ and ${\mathcal L}\{y\}$ and we can compute the inverse Laplace transforms separately. We start with ${\mathcal L}\{x\}$. $$\eqalign { \frac{s+9}{(s-6)(s+2)}&=\frac{A}{s-6}+\frac{B}{s+2} \cr s+9&=A(s+2)+B(s-6) \cr s+9&=(A+B)s+(2A-6B) \cr } $$ This gives us two equations in two unknowns $$ \leqalignno { 1&=A+B&(s) \cr 9&=2A-6B&(1) \cr} $$ which we solve to find $A=15/8$ and $B=-7/8$. So $$\eqalign { x(t)&={\mathcal L}^{-1}\{\frac{15/8}{s-6}-\frac{7/8}{s+2}\} \cr &=\frac{15}{8}e^{6t}-\frac{7}{8}e^{-2t} \cr } $$ Next we find $y(t)$. $$ \eqalign { \frac{2s-3}{s^2-4s-12}&=\frac{A}{s-6}+\frac{B}{s+2} \cr 2s-3&=A(s+2)+B(s-6) \cr 2s-3&=(A+B)s+(2A-6B) \cr } $$ This gives us two equations in two unknowns $$ \leqalignno { 2&=A+B&(s) \cr -3&=2A-6B&(1) \cr} $$ which we solve to find $A=9/8$ and $B=7/8$. So $$\eqalign { y(t)&={\mathcal L}^{-1}\{\frac{9/8}{s-6}+\frac{7/8}{s+2}\} \cr &=\frac{9}{8}e^{6t}+\frac{7}{8}e^{-2t} \cr } $$ There, that wasn't so bad was it? The only trick is that the algebra is a little messier. It is also possible to solve systems like these without Laplace transforms, but the usual techniques for doing so require more matrix theory than is included in the prerequisites for this class.

You can generate additional examples of first order linear systems of equations here.