Linear System Example: Distinct Real Roots

We solve Example 1 of §56 in [Simmons, Second edition], on p. 429,

\{ \begin{matrix}\frac{dx}{dt} & = & x + y \\ \frac{dy}{dt} & = & 4x − 2y \end{matrix}

by an alternate method. We apply the Laplace transform with the generic initial conditions x(0) = k_{1} and y(0) = k_{2} to get

\{ \begin{matrix}pX − k_{1} & = & X + Y \\ pY − k_{2} & = & 4X − 2Y \end{matrix} or \{ \begin{matrix}(1 − p)X + Y & = & −k_{1} \\ 4X + (−2 − p)Y & = & −k_{2}. \end{matrix}

The resemblance to (9) is not accidental. We solve these simultaneous linear equations for X and Y by elimination,

\{ \begin{matrix}[(−2 − p)(1 − p) − 1 · 4]X & = & −(−2 − p)k_{1} + k_{2} & = & pk_{1} + 2k_{1} + k_{2} \\ [(1 − p)(−2 − p) − 4 · 1]Y & = & −(1 − p)k_{2} + 4k_{1} & = & pk_{2} + 4k_{1} − k_{2}. \end{matrix}

The expressions in [ ] are both

(−2 − p)(1 − p) − 4 = p^{2} + p − 6 = (p + 3)(p − 2).

We therefore divide out by this and use (easy) partial fractions,

\{ \begin{matrix}X & = & \frac{pk_{1} + 2k_{1} + k_{2}}{(p + 3)(p − 2)} & = & \frac{4k_{1} + k_{2}}{5(p − 2)} + \frac{k_{1} − k_{2}}{5(p + 3)} \\ Y & = & \frac{pk_{2} + 4k_{1} − k_{2}}{(p + 3)(p − 2)} & = & \frac{4k_{1} + k_{2}}{5(p − 2)} + \frac{− 4k_{1} + 4k_{2}}{5(p + 3)} \end{matrix}

Finally, we apply the inverse Laplace transform and find

\{ \begin{matrix}x & = & \frac{4k_{1} + k_{2}}{5} e^{2t} + \frac{k_{1} − k_{2}}{5} e^{−3t} \\ y & = & \frac{4k_{1} + k_{2}}{5} e^{2t} − \frac{4(k_{1} − k_{2})}{5}e^{−3t}. \end{matrix}

This agrees with (12) in the book, if we take c_{1} = \frac{k_{1} − k_{2}}{5} and c_{2} = \frac{4k_{1} + k_{2}}{5} .

Compare this treatment with the traditional one in the book to decide which is shorter or easier or more efficient. Draw your own conclusions.

110.302 Differential Equations JMB File: lsysreal, Rev. A; 30 Aug 2001; Page 1