I am familiar with first-order differential equations and how to solve them, but only for the case w

protestommb

protestommb

Answered question

2022-06-16

I am familiar with first-order differential equations and how to solve them, but only for the case when there is a single unknown function.
For the following case there are two unknown functions and I'm struggling to determine where to even begin. The two equations:
(1) f ( t ) = A f ( t ) + B g ( t )
(2) g ( t ) = B g ( t ) + A f ( t )
Where f and g are functions of the variable t, and A & B are constants. The initial conditions are: f ( 0 ) = 0 and g ( 0 ) = 1.
I already have knowledge of the answer for f(t), which is:
(3) f ( t ) = B A + B ( 1 e ( A + B ) t )
I would really like to understand how to tackle questions of this type: where there are two unknown functions. I would attempt to show my working out but I don't know where to start.
Therefore, I would like to ask if anyone could please provide a hint or suggestion for me, something which I can work off. If necessary, I will update my post with original working. Thank you for your time.

Answer & Explanation

kpgt1z

kpgt1z

Beginner2022-06-17Added 23 answers

This is an example of a linear system:
( f g ) = ( a b c d ) ( f g ) ,
of F = M F . The idea is to look for a solution of the form F = v e λ t , where v is some (constant) vector. This gives λ v = M v , i.e. λ is an eigenvalue of M, and v the corresponding eigenvector.
So we have a bunch of solutions, each corresponding to an eigenvalue/vector of the matrix. As the DE is linear and homogenous, we can just take linear combinations of these to get the general solution. Specifying initial conditions gives you an exact solution.
As a concrete example, for your original DE we have
( f g ) = ( A B A B ) ( f g ) .
The eigenvalues of the matrix are 0 and ( A + B ), and the respective eigenvectors are ( B , A ) and ( 1 , 1 ) . So the general solution is
( f g ) = C ( B A ) e 0 t + D ( 1 1 ) e ( A + B ) t .
So using the initial conditions:
f ( 0 ) = 0 0 = C B D g ( 0 ) = 1 1 = C A + D
So C = 1 A + B and D = B A + B . This gives the desired solution.
Summer Bradford

Summer Bradford

Beginner2022-06-18Added 11 answers

You can also sum both equations and get:
f + g = 0 f + g = c
With initial conditions we get:
c = f ( 0 ) + g ( 0 ) = 1
Then use this in the first DE and you have a first order DE:
f = A f + B g = A f + B ( 1 f ) = ( A + 1 ) f + B
This equation is separable and can be easily solved.

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