Simultaneous Homogeneous Linear First Order ODEs/Examples/y' - 3y + 2z = 0, y' + 4y - z = 0

Theorem

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

\(\text {(1)}: \quad\)

\(\ds \dfrac {\d y} {\d x} - 3 y + 2 z\)

\(=\)

\(\ds 0\)

\(\text {(2)}: \quad\)

\(\ds \dfrac {\d x} {\d z} + 4 y - z\)

\(=\)

\(\ds 0\)

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

\(\ds y\)

\(=\)

\(\ds C_1 e^{5 x} + C_2 e^{-x}\)

\(\ds z\)

\(=\)

\(\ds -C_1 e^{5 x} + 2 C_2 e^{-x}\)

Proof

Using the technique of Solution to Simultaneous Homogeneous Linear First Order ODEs with Constant Coefficients, we calculate the roots of the quadratic equation:

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

where:

\(\ds a\)

\(=\)

\(\ds -3\)

\(\ds b\)

\(=\)

\(\ds 2\)

\(\ds c\)

\(=\)

\(\ds 4\)

\(\ds d\)

\(=\)

\(\ds -1\)

That is:

$\paren {k - 3} \paren {k - 1} - 8 = 0$

or:

$k^2 - 4 k - 5 = 0$

This has roots:

\(\ds k_1\)

\(=\)

\(\ds 5\)

\(\ds k_2\)

\(=\)

\(\ds -1\)

We also obtain:

\(\ds \paren {k - 3} A + 2 B\)

\(=\)

\(\ds 0\)

\(\ds 4 A + \paren {k - 1} B\)

\(=\)

\(\ds -1\)

When $k = 5$ we get that $A + B = 0$.

When $k = -1$ we get that $2 A - B = 0$.

This provides us with the solutions:

\(\ds y\)

\(=\)

\(\ds e^{5 x}\)

\(\ds z\)

\(=\)

\(\ds e^{-5 x}\)

or:

\(\ds y\)

\(=\)

\(\ds e^{-x}\)

\(\ds z\)

\(=\)

\(\ds 2 e^{-x}\)

From these, the general solution is constructed:

\(\ds y\)

\(=\)

\(\ds C_1 e^{5 x} + C_2 e^{-x}\)

\(\ds z\)

\(=\)

\(\ds -C_1 e^{5 x} + 2 C_2 e^{-x}\)

$\blacksquare$

Sources

• 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 3$. Equations of higher order and systems of first order equations: $\S 3.2$ First order systems: Example $2$