# Solution to Simultaneous Homogeneous Linear First Order ODEs with Constant Coefficients

## Theorem

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

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.