Help with a particular Ordinary Differential Equation

Question

I did the following problem but I am coming up with the wrong answer.

Problem:

Use Laplace transforms to solve the following system. All unknowns are fuctions of $x$. \begin{eqnarray*} w'' + y + z &=& -1 \\ w + y'' - z &=& 0 \\ -w' - y' + z'' &=& 0 \\ w(0) = 0, w'(0) = 1 &\text{,}& y(0) = 0 \\ y'(0)= 0, z(0) = -1 &\text{,}& z'(0) = 1 \\ \end{eqnarray*}

Answer:

\begin{eqnarray*} s^2W(s) - 1 + Y(s) + Z(s) &=& -\frac{1}{s} \,\,\,\,\,\,\,\,\,\, \text{ eq 1}\\ W(s) + s^2Y(s) - Z(s) &=& 0 \,\,\,\,\,\,\,\,\,\, \text{ eq 2}\\ -sW(s) - sY(s) + s^2Z(s) + s + 1 &=& 0 \,\,\,\,\,\,\,\,\,\, \text{ eq 3}\\ \end{eqnarray*} Now I solve eq 2 for $W(s)$. \begin{eqnarray*} W(s) &=& -s^2Y(s) + Z(s) \\ s^2( -s^2Y(s) + Z(s) ) - 1 + Y(s) + Z(s) &=& -\frac{1}{s} \\ -s^4Y(s) + s^2Z(s) - 1 + Y(s) + Z(s) &=& -\frac{1}{s} \\ (s^2 + 1)Z(s) + (-s^4+1)Y(s) &=& 1 - \frac{1}{s} \,\,\,\,\,\,\,\,\,\, \text{ eq 4}\\ \end{eqnarray*} \begin{eqnarray*} -s( -s^2Y(s) + Z(s) ) - sY(s) + s^2Z(s) + s - 1 &=& 0 \\ s^3Y(s) - sZ(s) - sY(s) + s^2Z(s) + s - 1 &=& 0 \\ (s^2 - s)Z(s) + s^3Y(s) - sY(s) + s - 1 &=& 0 \,\,\,\,\,\,\,\,\,\, \text{eq 5}\\ \end{eqnarray*} Now eq 4 and eq 5 represent a system of linear equations with two unknowns. I now solve eq 4 for $Z(s)$. \begin{eqnarray*} (s^2 + 1)Z(s) + (-s^4+1)Y(s) &=& \frac{s-1}{s} \\ (s^2 + 1)Z(s) &=& (s^4-1)Y(s) + \frac{s-1}{s} \\ (s^3 + s)Z(s) &=& (s^5-s)Y(s) + s-1 \\ Z(s) &=& \frac{(s^5-s)Y(s) + s-1}{s^3+s} \\ \end{eqnarray*} Now I subsitue into eq 5. \begin{eqnarray*} (s^2 - s)(\frac{(s^5-s)Y(s) + s-1}{s^3+s}) + s^3Y(s) - sY(s) + s - 1 &=& 0 \\ (s-1)(\frac{(s^5-s)Y(s) + s -1)}{s^2+1})+(s^2-s)Y(s) + s - 1 &=& 0 \\ (s-1)((s^5-s)Y(s) + s -1)+(s^2+1)(s^2-s)Y(s) + s(s^2+1) + s^2-1 &=& 0 \\ (s-1)((s^5-s)Y(s) + s -1)+(s^4-s^3+s^2-s)Y(s) + s^3+s + s^2-1 &=& 0 \\ (s-1)(s^5-s)Y(s) + s^2-s -s + 1 +(s^4-s^3+s^2-s)Y(s) + s^3+s + s^2-1 &=& 0 \\ (s-1)(s^5-s)Y(s) + (s^4-s^3+s^2-s)Y(s) + s^3+ 2s^2-s &=& 0 \\ (s^6-s^5-s^2+s)Y(s) + (s^4-s^3+s^2-s)Y(s) + s^3+ 2s^2-s &=& 0 \\ (s^6-s^5+s^4)Y(s) + s^3+ 2s^2-s &=& 0 \\ \end{eqnarray*} \begin{eqnarray*} Y(s) &=& \frac{ -s^3- 2s^2+s} { (s^6-s^5+s^4) } = \frac{ -s^2- 2s+1} { (s^5-s^4+s^3) } \\ \end{eqnarray*} The book's answer is: \begin{eqnarray*} y(x) &=& \cos x + 1 \\ \end{eqnarray*} Therefore I conclude I am wrong. I am hoping somebody can point out where I went wrong. Bob

Answer 1

Check your Eqs.1-3. Do you agree with this ? :

Answer 2

You posted three changes to my equations. I do agree with two of changes. Here is what I now think: \begin{eqnarray*} s^2W(s) - s + Y(s) + Z(s) &=& -\frac{1}{s} \,\,\,\, eq1 \\ W(s) + s^2Y(s) - Z(s) &=& 0 \,\,\,\,\,\,\, eq2 \\ -sW(s) - sY(s) + s^2Z(s) + s - 1 &=& 0 \,\,\,\,\, eq3 \end{eqnarray*} It is not $ -s - 1$ because $z(0) = -1$. Do you agree with these three equations?

Bob