Help solving with Partial Fractions...

[random88hero]

I have been working on this problem for two days now and still can't seem to come across the correct answer.

y' = 0.0011y(1900-y) , y(0) = 128

initially when I solved it regularly I ended up needing imaginary numbers because I needed e^(something) to come out negative, but I assumed that was incorrect

next I went on and tried partial fractions and came up with this...

dy (1/y + 1/(1900-y)) = 2.09 dx

moving along I had

ln(y) - ln(1900-y) = 2.09x + c

then

ln (y/(1900-y)) = 2.09x +c

here is where I come into some problems, when I try solving for c, and then y, i get some crazy decimal that are extremely large.... any help

 

[galactus]

I get \(\displaystyle C=\frac{443}{60800}\approx .007286\)

\(\displaystyle y=\Large\frac{1900e^{\frac{209x}{100}}}{e^{\frac{209x}{100}}+1900C}\)

 

[soroban]

Hello, random88hero!

\(\displaystyle \frac{dy}{dx} \;=\;0.0011y(1900-y),\;\; y(0) = 128\)

\(\displaystyle \text{Multiply by }10,\!000\!:\;\;10,\!000\frac{dy}{dx} \;=\;11y(1900-y) \quad\Rightarrow\quad 10,000\frac{dy}{y(1900-y)} \;=\;11\,dx\)


\(\displaystyle \text{Partial Fractions: }\;\frac{1}{y(1900-y)} \:=\:\frac{A}{y} + \frac{B}{1900-y} \quad\Rightarrow\quad A = \frac{1}{1900},\;\;B = \frac{1}{1900}\)

. . \(\displaystyle \text{Hence: }\;\frac{1}{y(1900-y)} \;=\;\frac{1}{1900}\left[\frac{1}{y} + \frac{1}{1900-y}\right]\)


\(\displaystyle \text{We have: }\;10,\!000\cdot\frac{1}{1900}\left(\frac{1}{y} + \frac{1}{1900-y}\right) \;=\;11\,dx\)

\(\displaystyle \text{Integrate: }\;\tfrac{100}{19}\int\left(\frac{1}{y} + \frac{1}{1999-y}\right)dy \;=\;11\int dx \quad\Rightarrow\quad \tfrac{100}{19}\bigg[\ln|y|\; \;- \ln|1900-y|\bigg] \;=\;11x \;+\; C_1\)

. . \(\displaystyle \ln|y| - \ln|1900-y| \;=\;\tfrac{209}{100}x + C_2 \quad\Rightarrow\quad \ln\left|\frac{y}{1900-y}\right| \:=\:\tfrac{209}{100}x + C_2\)



\(\displaystyle \text{When }x = 0,\:y = 128\)

. . \(\displaystyle \ln\left|\frac{128}{1900-128}\right| \;=\;\tfrac{209}{100}(0) + C \quad\Rightarrow\quad C \:=\:\ln\left(\tfrac{128}{1772}\right) \;=\;\ln\left(\tfrac{32}{443}\right)\)


\(\displaystyle \text{Hence, we have: }\;\ln\left|\frac{y}{1900-y}\right| \;=\;\tfrac{209}{100}x + \ln\left(\tfrac{32}{443}\right)\)


If you want to solve for \(\displaystyle y\), go ahead . . . I'll wait in the car.