Mathematics: Consider the following initial value problem 5 points (1 + t) y’ = t y (4 – y), y(0) = y0 0. a) solve the initial value problem and find its general solution. b) determine how the solution behaves as t ø¶. c) if the initial value y0 = 2, find the time t when the solution first reaches the value 3.99.

The answer is BStep-by-step explanation:Ed2020...

Consider the following initial value problem 5

rosieposie27

05.01.2022

a) To determine its general solution we have to accommodate the differential equation:

$\frac{dy}{(4-y)y} = \frac{t*dt}{t+1}$

We proceed to integrate each part as follow

$\frac{1}{4} \int\limits {\frac{1}{y}+\frac{1}{4-y}} \, dy = \int\limits {1-\frac{1}{t+1} } \, dt$

Then we have:

$\frac{Ln(\frac{y}{y-4} )}{4} = t-Ln(t+1)+C\\\\Ln(\frac{y*(t+1)^{4}}{y-4} ) = 4t+4C\\\\\frac{y*(t+1)^{4}}{y-4} = e^{4t+4C} = K*e^{4t}$

Finally we have:

$y = \frac{4Ke^{4t} }{Ke^{4t}-(t+1)^{4}}$

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 Ø¶ "

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:

$3.99 = \frac{4e^{4t} }{e^{4t}+(t+1)^{4}}$

Finally, resolving the equation we have as a result:

t = 2.84

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