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?
\(\displaystyle \L\\ y'\, =\, y(4\, -\, y)\)
\(\displaystyle \L\\ y(t)\, =\, \frac{4}{1\, +\, Ce^{-4^t}}\)
*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?