Find exact solution to dy/dt = (y - 1)(3 - y), y(0) = 5

warsatan

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Sep 12, 2005
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36
dy/dt = (y-1)(3-y)
find exact solution with inital condition y(0) = 5

I worked this problem using partial fraction and got

0.5 ln (y-1) + 0.5ln (3-y) = t + c

i tried to plug in that initial condition and its getting no where, any help would be awesome, thanks
 
dy/dt = [y-1] / [ 3-y] rewrite
{ -[y-3] / [y-1]} dy=dt but it is a improper fraction , divide
1
_____
y-1 l y-3
y-1
-------
-2

or we have -[1 -2/[y-1]} dy = dt
[2dy/[y-1] ] - dy = dt integrate
2 ln[y-1] -y =t + c at t=0 y=5
2 ln[4] -5 =c

2ln[y-1] -y =t + 2 ln4 -5 answer

please check for errors
 
Separate variables and integrate:

\(\displaystyle \L\\\int\frac{1}{(y-1)(3-y)}=\int{dt}\)

\(\displaystyle \L\\\frac{ln(y-1)-ln(y-3)}{2}=t+C\)

Solve for y:

\(\displaystyle \L\\y=\frac{3e^{2t+2C}-1}{e^{2t+2C}-1}\)

Initial condition, solve for C:

\(\displaystyle \L\\5=\frac{3e^{2(0)+2C} \;\ -1}{e^{2(0)+2C} \;\ -1}\)

\(\displaystyle \L\\C=\frac{ln(2)}{2}\)
 
galactus said:
Separate variables and integrate:

\(\displaystyle \L\\\int\frac{1}{(y-1)(3-y)}=\int{dt}\)

\(\displaystyle \L\\\frac{ln(y-1)-ln(y-3)}{2}=t+C\)

Solve for y:

\(\displaystyle \L\\y=\frac{3e^{2t+2C}-1}{e^{2t+2C}-1}\)

Initial condition, solve for C:

\(\displaystyle \L\\5=\frac{3e^{2(0)+2C} \;\ -1}{e^{2(0)+2C} \;\ -1}\)

\(\displaystyle \L\\C=\frac{ln(2)}{2}\)

as always, thank you galactus
 
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