Systems of Differential Equations In the previous videos we covered the various
ways that differential equations were classified; another classification of differential equations
depends on the number of unknown functions that are involved. If there is a single function
to be determined, then one equation is sufficient. However, if there are two or more unknown
functions, then a system of equations is required. Up to this point we have been discussing single
differential equations containing one unknown function. But often in theory, as well as
in many applications, we must deal with systems of differential equations.
A system of ordinary differential equations is two or more equations involving the derivatives
of two or more unknown functions of a single independent variable.
For example if x(t) and y(t) denote dependent variables and t denotes the independent variable,
then a system of two first-order differential equations is given by following expressions:
A solution of this system is a pair of differentiable functions, that satisfies each equation in
the system on some common interval I. This is analogous to systems of linear equations
from your studies of algebra, recall that the goal was to find an ordered pair x and
y that would satisfy both equations. In the case of a system of differential equations
we are actually looking for two functions that would satisfy both differential equations.
Systems of differential equations will be discussed in greater detail in a much later
video. For now this video will illustrate how to identify and verify a solution to the
system of differential equations. Let’s go over a couple of examples to illustrate
how to verify the solutions of a system of differential equations.
Verify that the indicated pair of functions is a solution of the given system of differential
equations on the interval negative infinity to positive infinity, or all real numbers.
The system of differential equations is the derivative of x with respect to t equals x
plus 3y and the derivative of y with respect to t equals 5x plus 3y. The solutions to verify
are x = e^-2t + 3e^6t and y = -e^-2t + 5e^6t. Notice that we have two dependent variables,
specifically, x and y both of which contain a single independent variable in this case
t. The procedure to verify a system of differential equations is very similar to the way that
you verified ordinary differential equations. It is just a matter of finding all the required
derivatives and substituting them, along with the functions, into the system of differential
equations, making sure that the left and right hand side of each equation match.
In this example we need to find the derivative of x with respect to time, so we go ahead and
apply the chain rule to the first function, taking the derivative term by term results
in the following expression for the derivative. Now let’s find the derivative of y with
respect to t, similar to the previous derivative we need to find the derivative term by term
and apply the chain rule, this results in the following expression.
Having found the derivative we go ahead and verify that both the left and right hand side
of each differential equation matches when we substitute the derivative and functions.
Let’s verify the first differential equation. We start by replacing the left side with the
derivative and replace the right side with the functions x and y as follows. Then it
is just a matter of simplifying the expressions so we go ahead and distribute the constant
on the right side, then we collect like terms, in the end we see that both sides of the equation
match. Now we need to do the same for the second differential equation. So we go ahead
and substitute the derivative and the functions with their respective expressions, then we
go ahead and distribute the constants on the right side of the equation, and finally we
collect like terms. In the end we see that both sides also match. Since these functions
are solutions to both differential equations we conclude that these two functions are indeed
solutions to the system of differential equations.
The process is very similar to the way we verified solutions to differential equations,
in this case we need to verify all the equations within the system. Alright lets go over the
final example. The system of differential equations is the
second derivative of x with respect to t equals 4y plus e^t and the second derivative of y
with respect to t equals 4x minus e^t. The solutions to verify are x = cos(2t) + sin(2t)
+ (1/5)*e^t and y = -cos(2t) - sin(2t) - (1/5)*e^t. Notice that we once again have two dependent
variables in this case x and y both of which contain the independent variable t.
Alright similar to the 1st example let's first find the derivatives of the dependent variables
with respect to t. Notice that we need to find the second derivative of each function
so let’s go ahead and find the first derivative of x with respect to t. Taking the derivative
term by term along with the chain rule we obtain the following expression. Next let’s
take the derivative once more, taking the derivative a second time results in the following expression.
Next let’s do the same and find the second derivative of y with respect to t.
Taking the derivative term by term and applying the chain rule we obtain the following expression,
then we once again take the derivative with respect to t. Doing that we obtain the
following expression for the second derivative. Now it’s just a matter of substituting the
second derivatives along with the functions and verify that they satisfy the system of
differential equations. Let's verify the first differential equation by substituting these
expressions and verifying that the left and right hand side of the equation match. Substituting
the expressions we obtain the following, then we go ahead and distribute the constant on
the right side of the equation, from here we go ahead and collect like terms, in the
end we see that both sides of the equation match. Now we need to verify second differential
equation, so we go ahead and substitute the following expressions into the differential equation,
then we go ahead and distribute the constant on the right side of the equation,
from here we collect like terms, in the end we see that both sides of the equation match.
Since these functions are solutions to both differential equations we conclude that these
two functions are indeed solutions to the system of differential equations.
Alright in our next video we will start discussing initial value problems. Also known as IVP’s