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In the introduction of this section we briefly talked how a system of differential equations can occur from a population problem wherein we remain track of the population of both the prey and the predator. This makes sense that the number of prey present will influence the number of the predator present. Similarly, the number of predator present will influence the number of prey present. Thus the differential equation which governs the population of either the prey or the predator must in some way based on the population of the other. It will lead to two differential equations which must be solved simultaneously so as to determine the population of the predator and the prey.
The entire point of this is to see that systems of differential equations can occur quite simple from naturally occurring situations. Developing an effectual predator-prey system of differential equations is not the subject of this section. Though, systems can occur from nth order linear differential equations suitably. Before we find this though, let's write down a system and find some terminology out of the way.
We are going to be searching at first order, linear systems of differential equations. These terms implies the same thing which they have meant up to this point. The main derivative anywhere in the system will be a first derivative and each unknown function and their derivatives will only arise to the first power and will not be multiplied with other unknown functions. Now there is an example of a system of first order, linear differential equations.
x1' = x1 + 2x2
x2' = 3x1 + 2x2
We call this type of system a coupled system as knowledge of x2 is needed in order to get x1 and similarly knowledge of x1 is needed to get x2. We will worry regarding that how to go about solving these presently. At this point we are only involved in becoming familiar along with some of the fundamentals of systems.
Here, as mentioned earlier, we can write an nth order linear differential equation like a system. Let's notice how that can be done.
If the roots of the equation (a-b) x 2 + (b-c) x+ (c - a)= 0 are equal. Prove that 2a=b+c. Ans: (a-b) x 2 + (b-c) x+ (c - a) = 0 T.P 2a = b + c B 2 - 4AC = 0
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