Differential Equation $y' = \frac{ty(4-y)}{(1+y)}$

Question

$y' = \frac{ty(4-y)}{(1+y)}$

given $ y_0 = 2 $ At what time $t$ when the solution will first be 3.9

I tried solving this but it didnt work out so well.

What I did was:

1. separate dy/dx and move all $y$ terms to one side with $dy$
2. Some how I don't know how to deal with the $dx$, so I couldn't come up with the solution , $y$.

Answer 1

Separate the $y$ and $t$ terms on either side of the equal side and get

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

The right-hand side may be broken up as

$$\begin{align} \frac{1+y}{y (4-y)} &= \frac{1}{y (4-y)} + \frac{1}{4-y} \\ &=\frac{1}{4} \left ( \frac{1}{y} + \frac{1}{4-y} \right ) + \frac{1}{4-y} \\ &= \frac{1}{4 y} + \frac{5}{4} \frac{1}{4-y} \end{align}$$

Now integrate both sides:

$$ \frac{1}{2} t^2 + C = \frac{1}{4} [\log{y} - 5 \log{(4-y)}] $$

or

$$ 2 t^2 + C = \log{\left [ \frac{y}{(4-y)^5} \right ]}$$

Using $y(0)=2$ implies that $C=-\log{16}$. Then

$$t^2 = \frac{1}{2} \log{\left [ \frac{16 y}{(4-y)^5} \right ]}$$

When $y=3.9$, $t$ will be

$$t^2 = \frac{1}{2} \log{\left [ \frac{62.4}{(0.1)^5} \right ]}$$

This gives

$$t \approx 2.797$$

Answer 2

Here, $y'\neq \frac{dy}{dx}$ but $\frac{dy}{dt}$ so, the differential equation after rearrangement becomes,

$$\frac{(1+y)}{y(4-y)}dy=tdt$$

Answer 3

You need to use $ \frac {dy}{dt} = y' $. Thus, after re-arrangements; you obtain:

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

Integrating both sides, you get:

$$ \ln \left( \frac{y}{(4 - y)^5}\right) = t^2 + K $$

Apply the limit conditions here, and you get the result.