### Differential equations: Systems of differential equations

### Systems of coupled linear first-order ODEs

For a brief look at more than one differential equation with several unknown functions, we consider the following **system of coupled homogeneous linear first-order differential equations:** \[\eqs{\dfrac{\dd x}{\dd t} &= &a\cdot x+ b\cdot y\cr \dfrac{\dd y}{\dd t} &=& c\cdot x + d\cdot y}\] where #x# and #y# are unknown functions of #t#, and #a#, #b#, #c#, #d# are constants. This system can be solved by use of the theory of homogeneous linear second-order ODEs with constant coefficients.

Conversion of two coupled homogeneous linear first-order differential equations to a homogeneous linear second-order equation

Let #a#, #b#, #c#, #d#, #p#, #q# be constants, and let #x#, #y#, #z# be functions of #t#.

- If #\rv{x,y}# is a solution of the system of two coupled first order differential equations homogeneous linear \[ \eqs{x' &= &a\cdot x+ b\cdot y\cr y' &=& c\cdot x + d\cdot y}\] then both #x# and #y# are solutions of the homogeneous linear second order differential equation \[ z''-(a+d)\cdot z'+(a\cdot d-b\cdot c)\cdot z =0\]
- If #x# and #y# are linearly independent solutions of the linear second-order differential equation \[z''+p\cdot z'+q\cdot z =0\] then there are constants #a_1#, #b_1#, #c_1#, #d_1# such that \[\eqs{x' &= &a_1\cdot x+ b_1\cdot y\cr y' &=& c_1\cdot x + d_1\cdot y}\]

We know that the type of solution of the above second-order ODE depends on the discriminant of the corresponding characteristic polynomial \[\lambda^2 -(a+d)\lambda+(ad-bc)\] This discriminant \(D\) can be rewritten in terms of the system of coupled first-order differential equations: \[D=(a+d)^2-4(a\cdot d-b\cdot c)=(a-d)^2+4b\cdot c\] We now distinguish again three cases (statements regarding #x(t)# also hold for #y(t)#):

- If \((a-d)^2+4b\cdot c\gt0\) then there are two real solutions of the characteristic equation: \[ \lambda_{1,2}=\frac{a+d\pm\sqrt{(a-d)^2+4b\cdot c}}{2}\] and the general solution for \(x(t)\) is \[x(t)=c_1\cdot \e^ {\lambda_1t}+c_2\cdot \e^ {\lambda_2t}\]
- If \((a-d)^2+4b\cdot c=0\) then there is a single real solution of the characteristic equation: \[ \lambda=\frac{a+d}{2}\]and the general solution for \(x(t)\) is \[x(t)=(c_1+c_2 t)\cdot \e^ {\lambda t}\]
- If \((a-d)^2+4b\cdot c\lt0\) then there are no real solutions and there are two complex solutions of the characteristic equation: \[ \lambda_{1,2}=\alpha\pm{\ii}\,\beta\] with \[\alpha=\frac{a+d}{2}\quad\text{and}\quad\beta=\frac{\sqrt{-(a-d)^2-4b\cdot c}}{2}\] and the general solution for \(x(t)\) is \[x(t)=\e^ {\alpha t}\cdot\bigl(c_1\cdot \cos(\beta t)+c_2\cdot \sin(\beta t)\bigr)\]

We show how the first statement of the theorem can be used to solve the coupled system.

Solution of the coupled system of first-order equations by using the second-order equation

The general solution of the coupled system of differential equations \[\eqs{x' &= &a\cdot x+ b\cdot y\cr y' &=& c\cdot x + d\cdot y}\]

can be found as follows:

- Solve the linear second-order equation \(z''-(a+d)\cdot z'+(a\cdot d-b\cdot c)\cdot z =0\); this provides a pair of linearly independent solutions #u# and #v# of that equation.
- Consequently, solutions #x# (and #y#) of the system have the form #x=A\cdot u+B\cdot v# (and #y=C\cdot u+D\cdot v#, respectively), for yet to be determined constants #A#, #B#, #C#, #D#. Substitute these expressions for #x# and #y# in the system.
- The result of the previous step is a pair of ordinary linear equations in the unknowns #A#, #B#, #C#, and #D#. Use these equations to express two of them in the other two.
- Substitute the expressions for two of the four constants found in the previous step in the equations of step 2 in order to find the general form for #x# and for #y#.

According to the

*Conversion of two coupled homogeneous linear first-order differential equations to a homogeneous linear second-order equation*, the coordinates #\rv{x,y}# of a solution of this system are also solutions of the linear second-order ODE \( z''-(1+2) z'+ (1\cdot2-2\cdot3) z = 0 \), which can be simplified to

\[z''-3 z'-4 z = 0\]

The

*characteristic equation*#\lambda^2-3\lambda -4 = 0# has solution #\lambda=4\lor\lambda=-1#, so the general solution of this second-order equation is

\[z=A\cdot \euler^{4\cdot t}+B\cdot \euler^ {- t }\]

Because #x# and #y# are also solutions, we can write

\[ \eqs{x &= &A\cdot \euler^{4\cdot t}+B\cdot \euler^ {- t } \cr y &=&C\cdot \euler^{4\cdot t}+D\cdot \euler^ {- t }\cr}\]

for numbers #A#, #B#, #C#, #D#, that are yet to be determined.

If we fill these expressions for #x# and #y# into the coupled system, then we find a system of linear equations from which the following constants to be determined can be found:

\[\begin{array}{rcl}

\dfrac{\dd}{\dd t}\left(A\cdot \euler^{4\cdot t}+B\cdot \euler^ {- t }\right) &=&

\left(A\cdot \euler^{4\cdot t}+B\cdot \euler^ {- t }\right)

+ 2 \left( C\cdot \euler^{4\cdot t}+D\cdot \euler^ {- t }\right)\\

\dfrac{\dd}{\dd t}\left(C\cdot \euler^{4\cdot t}+D\cdot \euler^ {- t }\right) &=&

3 \left(A\cdot \euler^{4\cdot t}+B\cdot \euler^ {- t }\right)

+2 \left( C\cdot \euler^{4\cdot t}+D\cdot \euler^ {- t } \right)\\

&&\phantom{xx}\color{blue}{\text{the solutions found}}\\

4\cdot A\cdot \euler^{4\cdot t}-B\cdot \euler^ {- t }&=&2\cdot C\cdot \euler^{4\cdot t}+A\cdot \euler^{4\cdot t}+2\cdot D\cdot \euler^ {- t }+B\cdot \euler^ {- t } \\ 4\cdot C\cdot \euler^{4\cdot t}-D\cdot \euler^ {- t }&=&2\cdot C\cdot \euler^{4\cdot t}+3\cdot A\cdot \euler^{4\cdot t}+2\cdot D\cdot \euler^ {- t }+3\cdot B\cdot \euler^ {- t } \\

&&\phantom{xx}\color{blue}{\text{derivative calculated and brackets cleared}}\\

4\cdot A&=&2\cdot C+A \\ 4\cdot C&=&2\cdot C+3\cdot A \\

-B&=&2\cdot D+B \\ -D&=&2\cdot D+3\cdot B \\

&&\phantom{xx}\color{blue}{\text{the coefficients of }\euler^{4\cdot t}\text{ and }\euler^ {- t }\text{ compared}}\\

A&=&\displaystyle {{2\cdot C}\over{3}} \\ B&=&\displaystyle -D \\

&&\phantom{xx}\color{blue}{\text{linear equations in }A,B,C,D\text{ solved}}\\

x&=&\displaystyle {{2\cdot C\cdot \euler^{4\cdot t}}\over{3}}-D\cdot \euler^ {- t } \\ y&=&\displaystyle C\cdot \euler^{4\cdot t}+D\cdot \euler^ {- t } \\

&&\phantom{xx}\color{blue}{\text{solution substituted into function rules for }x\text{ and }y}\\

\end{array}\]

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