Answer:
a) To determine its general solution we have to accommodate the differential equation:
[tex]\frac{dy}{(4-y)y} = \frac{t*dt}{t+1}[/tex]
We proceed to integrate each part as follow
[tex]\frac{1}{4} \int\limits {\frac{1}{y}+\frac{1}{4-y}} \, dy = \int\limits {1-\frac{1}{t+1} } \, dt[/tex]
Then we have:
[tex]\frac{Ln(\frac{y}{y-4} )}{4} = t-Ln(t+1)+C\\\\Ln(\frac{y*(t+1)^{4}}{y-4} ) = 4t+4C\\\\\frac{y*(t+1)^{4}}{y-4} = e^{4t+4C} = K*e^{4t}[/tex]
Finally we have:
[tex]y = \frac{4Ke^{4t} }{Ke^{4t}-(t+1)^{4}}[/tex]
Replacing the initial value y(0) = y0, we have:
K = y0/(y0-4)
b) Please, could you describe better what do you need because I don't understand this "t ض "
c) So we first need to find K, we replace y0=2 and we have:
K = -1
then the equation, replacing y=3.99 is:
[tex]3.99 = \frac{4e^{4t} }{e^{4t}+(t+1)^{4}}[/tex]
Finally, resolving the equation we have as a result:
t = 2.84
