I have a first order linear differential equation (a variation on a draining mixing tank problem) wi

Cristopher Knox 2022-07-10 Answered
I have a first order linear differential equation (a variation on a draining mixing tank problem) with many constants, and want to separate variables to solve it.
d y d t = k 1 + k 2 y k 3 + k 4 t
y is the amount of mass in the tank at time t, and for simplicity, I've reduced various terms to constants, k 1 through k 4 .
Separation of variables is made difficult by k 1 , and I've considered an integrating factor, but think I might be missing something simple.
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Answers (1)

Alisa Jacobs
Answered 2022-07-11 Author has 13 answers
d y d t = k 1 + k 2 y k 3 + k 4 t
For simplicity substitute u = k 3 + k 4 t
k 4 d y d u = k 1 + k 2 y u
y k 2 k 4 u y = k 1 k 4
Solve the homogeneous DE and use variation of parameter method for the inhomogeneous DE:
y k 2 k 4 u y = 0

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I am solving a differential equation d y d t = y + 1 t + 1 .
I got the solution y = c ( t + 1 ) 1, c a constant.
But the handout by my professor says
"The solution is y = c ( t + 1 ) 1, c a constant. where t 1. But it does not mean that y is not defined at t 1. It means that y can take any value at t = 1 as long as it satisfies y = c ( t + 1 ) 1."
It does not make much sense to me because if y satisfies the equation y = c ( t + 1 ) 1 at t = 1, then the solution will be just y = c ( t + 1 ) 1, c a constant.
And I believe my solution indeed makes sense because y = c ( t + 1 ) 1, c a constant is defined on R and differentiable, y = c for any t R . On the other hand, plugging this in to y + 1 t + 1 gives me c for any t R .
Would you explain what the handout is saying?

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