Given a function f with \(\displaystyle{f{{\left({1},{940}\right)}}}={1},{000},{000}\), and

Kason Palmer

Kason Palmer

Answered question

2022-03-25

Given a function f with f(1,940)=1,000,000, and its derivative is equal to kf(200f) and k is a constant. How can I get f?

Answer & Explanation

aznluck4u72x4

aznluck4u72x4

Beginner2022-03-26Added 16 answers

Solution.
If you had dydx=y(5y), you'd probably want to treat it as a separable equation:
dyy(5y)=dx.
Use partial fractions on the left and integrate to get something in terms of log's. What you did was integrate with respect to f, not x, when the derivative was with respect to x.
Ireland Vaughan

Ireland Vaughan

Beginner2022-03-27Added 14 answers

Step 1
Well, if we have the DE:
y'(x)=k·y(x)·(n-y(x))       (1)
We can divide both sides by y(x)(ny(x)), in order to write:
y'(x)y(x)·(n-y(x))=k      (2)
Integrate both sides with respect to x:
y'(x)y(x)·(n-y(x)) dx=k dx     (3)
And it is really not that hard to find:
1n·lny(x)y(x)-n=kx+C      (4)
Step 2
Now, for your case we can solve for C using the initial condition:
1200·ln106106-200=k·1940+CC=1200·ln(50004999)-1940k     (5)
So: 1200·lny(x)y(x)-200=k(x-1940)+1200·ln(50004999)      (6)

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