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

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aznluck4u72x4 · Accepted Answer

Solution.If you had dydx=y(5−y), you&#039;d probably want to treat it as a separable equation:dyy(5−y)=dx.Use partial fractions on the left and integrate to get something in terms of log&#039;s. What you did was integrate with respect to f, not x, when the derivative was with respect to x.

Ireland Vaughan · Accepted Answer

Step 1Well, if we have the DE:y&#039;(x)=k·y(x)·(n-y(x))&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(1)We can divide both sides by&amp;nbsp;y(x)⋅(n−y(x)), in order to write:y&#039;(x)y(x)·(n-y(x))=k&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(2)Integrate both sides with respect to x:∫y&#039;(x)y(x)·(n-y(x))&amp;nbsp;dx=∫k&amp;nbsp;dx&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(3)And it is really not that hard to find:1n·lny(x)y(x)-n=kx+C&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(4)Step 2Now, for your case we can solve for C using the initial condition:1200·ln106106-200=k·1940+C⟺C=1200·ln(50004999)-1940k&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(5)So:&amp;nbsp;1200·lny(x)y(x)-200=k(x-1940)+1200·ln(50004999)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(6)

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

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