Solution to Simultaneous Homogeneous Linear First Order ODEs with Constant Coefficients

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Theorem

Consider the system of linear first order ordinary differential equations with constant coefficients:

\(\text {(1)}: \quad\) \(\ds \dfrac {\d y} {\d x} + a y + b z\) \(=\) \(\ds 0\)
\(\text {(2)}: \quad\) \(\ds \dfrac {\d x} {\d z} + c y + d z\) \(=\) \(\ds 0\)


The general solution to $(1)$ and $(2)$ consists of the linear combinations of the following:

\(\ds y\) \(=\) \(\ds A_1 e^{k_1 x}\)
\(\ds z\) \(=\) \(\ds B_1 e^{k_1 x}\)

and:

\(\ds y\) \(=\) \(\ds A_2 e^{k_2 x}\)
\(\ds z\) \(=\) \(\ds B_2 e^{k_2 x}\)

where $A_1 : B_1 = A_2 : B_2 = r$

where $r$ is either of the roots of the quadratic equation:

$\paren {k + a} \paren {k + d} - b c = 0$


Proof

We look for solutions to $(1)$ and $(2)$ of the form:

\(\text {(3)}: \quad\) \(\ds y\) \(=\) \(\ds A e^{k x}\)
\(\text {(4)}: \quad\) \(\ds z\) \(=\) \(\ds B e^{k x}\)

We do of course have the Trivial Solution of Homogeneous Linear 1st Order ODE:

$y = z = 0$

which happens when $A = B = 0$.

So let us investigate solutions where either or both of $A$ and $B$ are non-zero.


Substituting $(3)$ and $(4)$ into $(1)$ and $(2)$ and cancelling $e^{k x}$, we get::

\(\text {(5)}: \quad\) \(\ds \paren {k + a} A + b B\) \(=\) \(\ds 0\)
\(\text {(6)}: \quad\) \(\ds c A + \paren {k + d} B\) \(=\) \(\ds 0\)


From $(5)$ and $(6)$ we get:

\(\text {(7)}: \quad\) \(\ds \) \(\) \(\ds \paren {\paren {k + a} \paren {k + d} - b c} A\)
\(\ds \) \(=\) \(\ds \paren {\paren {k + a} \paren {k + d} - b c} B\)
\(\ds \) \(=\) \(\ds 0\)

So $A = B = 0$ unless $k$ is a root of the quadratic equation:

$\paren {k + a} \paren {k + d} - b c = 0$

That is:

$(8): \quad \begin {vmatrix} k + a & b \\ c & k + d \end {vmatrix} = 0$

where the above notation denotes the determinant.


Assume $(8)$ has distinct roots $k_1$ and $k_2$.

Taking $k = k_1$ and $k = k_2$ in $(7)$, we can obtain ratios $A_1 : B_1$ and $A_2 : B_2$ such that:

\(\ds y\) \(=\) \(\ds A_1 e^{k_1 x}\)
\(\ds z\) \(=\) \(\ds B_1 e^{k_1 x}\)

and:

\(\ds y\) \(=\) \(\ds A_2 e^{k_2 x}\)
\(\ds z\) \(=\) \(\ds B_2 e^{k_2 x}\)

are solutions of $(1)$ and $(2)$.

By taking arbitrary linear combinations of these, we obtain the general solution.


This needs considerable tedious hard slog to complete it.
In particular: Cover the case where $k_1 = k_2$. The source work is vague on this subject. Recommend this solution be reworked, preferably in conjunction with a more rigorous and thorough source work than the one used here.
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Sources

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