Show that y = 4/(1 + Ce^(-4t)) solves y' = y(4 - y)

mathstresser

Junior Member
Joined
Jan 28, 2006
Messages
134
Show that the given solution is a general solution of the differential equation.

\(\displaystyle \L\\ y'\, =\, y(4\, -\, y)\)

\(\displaystyle \L\\ y(t)\, =\, \frac{4}{1\, +\, Ce^{-4^t}}\)

*note: The last part is e to the power -4t
the first equation is y prime=...

C = 1, 2, ..., 5

\(\displaystyle \L\\ dy/dx\, =\, y(4\, -\, y)\)

\(\displaystyle \L\\ \frac{dy}{4y\, -\, y^2}\, =\, dx\)

\(\displaystyle \L\\ \int\, \frac{dy}{4y\, -\, y^2}\, =\, \int\, dx\)

The second integral is, of course, x. But what is the first integral?
 
I would try using partial fractions to integrate.

Eliz.
 
Re: diff equations and solutions

Hello, mathstresser!

Show that the given solution is a general solution of the differential equation.

Differential Equation: \(\displaystyle \L\;y' =\: y(4\,-\, y)\)

Solution: \(\displaystyle \L\;y(t)\:=\:\frac{4}{1\, +\, Ce^{-4t}}\)

"Show that ..." means just plug it in.

If they said: "Show that \(\displaystyle x\,=\,6\) is a solution of: \(\displaystyle \,x^4\,-\,7x^3\,+\,5x^2\,+\,8x\,-\,12\:=\:0\)"
. . they don't expect you to solve the equation . . . right?


The differential equation is: \(\displaystyle \L\:y'(t) \:=\:y(4\,-\,y)\)

The left side is: \(\displaystyle \L \:\fbox{y'(t)\:=\:\frac{16e^{-4t}}{(1\,+\,Ce^{-4t})^2}}\)

The right side is: \(\displaystyle \L\:y(4\,-\,y)\:=\:\frac{4}{1\,+\,Ce^{-4t}}\left(4\,-\,\frac{4}{1\,+\,Ce^{-4t}}\right)\)

Are these two expression equal?
Let's simplify the right side . . .

We have: \(\displaystyle \L\:\frac{4}{1\,+\,Ce^{-4t}}\left(\frac{4(1\,+\,Ce^{-4t})\,-\,4}{1\,+\,Ce^{-4t}}\right) \;=\;\frac{4}{1\,+\,Ce^{-4t}}\,\cdot\,\frac{4Ce^{-4t}}{1\,+\,Ce^{-4t}}\)

. . . \(\displaystyle \L= \;\fbox{\frac{16Ce^{-4t}}{(1\,+\,Ce^{-4t})^2}}\)


Yes, they are equal . . . The solution is correct.

 
I understand the right side of the equation, but not the left side. How do you get that?
 
mathstresser said:
I understand the right side of the equation, but not the left side. How do you get that?
By differentiating the given y(t) to find y'(t).

Eliz.
 
Top