I was thinking of functions who have differential equations in which the function is 'factorable' fr

ahiahysel

ahiahysel

Answered question

2022-02-24

I was thinking of functions who have differential equations in which the function is 'factorable' from first order differential equation. Eg: Functions satisfying the equation:
dydx=η(x)y(x)
Example functions:
y=eax,y=11x.. etc
Now, I notice that the solution to the differential equation is:
y=eηdx
But here is the interesting thing, the above expression has the same form of the function which is used when we solve linear differential equation

Answer & Explanation

Mikayla Swan

Mikayla Swan

Beginner2022-02-25Added 9 answers

dydx=η(x)y(x)
First, change of symbol x for t:
dydt=η(t)y(t)
dydtη(t)y(t)=0
Second, let η(t)=p(t)
dydt+p(t)y(t)=0
This is the equation from your cited document in the particular case g(t)=0:
dydt+p(t)y(t)=g(t)
So it is not surprizing that you observe the relationship of solutions.

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