Solve the following set of equations using Laplace Transforms

Question

The two equations are $$y''+z''-z'=0$$ $$y'+z'-2z = 1-e^{-t}$$ where $y_0 =0 , z_0 = 1$, $y'_0\text{ and } z'_0 = 1$ I've found that $$L(y') = pL(y)-y_0$$ $$L(y'') = p^2L(y)-py_0-y'_0$$ $$L(1-e^{-t}) = \frac{1}{p}-\frac{1}{p+1}$$ Which gives the set of equations $$p^2L(y)-1+p^2L(z)-p-1-pL(z)+1=0$$ $$pL(y)+pL(z)-1-2L(z)=\frac{1}{p}-\frac{1}{p+1}$$ From here I'm unsure how to solve for $y$ and $z$. How do I calculate the inverse Laplace transform after solving this set of equations? Any help is appreciated.

Answer 1

You have the following system of equations in $L(y)$ and $L(z)$: \begin{eqnarray} pL(y)+(p-1)L(z)&=&\dfrac{1}{p}+1\\ pL(y)+(p-2)L(z)&=&\dfrac{1}{p}-\dfrac{1}{p+1}+1 \end{eqnarray} Subtracting the first and second equations we get $$ L(z)=\dfrac{1}{p+1}, $$ then substituting $L(z)$ into the first equation we get \begin{eqnarray} pL(y)+\dfrac{p-1}{p+1}&=&\dfrac{1}{p}+1\\ pL(y)&=&\dfrac{1}{p}+1-\dfrac{p-1}{p+1}\\ pL(y)&=&\dfrac{1}{p}+\dfrac{2}{p+1}\\ L(y)&=&\dfrac{1}{p^2}+\dfrac{2}{p(p+1)}\\ L(y)&=&\dfrac{1}{p^2}+\dfrac{2}{p}-\dfrac{2}{p+1}. \end{eqnarray} Hence \begin{eqnarray} y(t)&=&tu(t)+2u(t)-2e^{-t}u(t)=(t+2-2e^{-t})u(t)\\ z(t)&=&e^{-t}u(t) \end{eqnarray} with $u$ the unit step function, i.e. $$ u(t)=\begin{cases} 0 & \text{ if } t<0\\ 1 & \text{ if } t \ge 0 \end{cases}. $$