Consider the initial value problem $$y'=ty(4-y)/(1+t)$$ $$y(0)=y_{0}>0$$

(a)Determine how the solution behaves as $t$ tends to infinity.

(b)If $y_{0}=2$,find the time $T$ at which the solution first reaches the value of 3.99

(c)Find the range of initial values for which the solution lies in the interval $3.99<y<4.01$ by the time $t=2$.

What i tried

(a)I first solve the following IVP to get a general solution of $$y=\frac{4A}{(1+t)^{4}e^{-4t}+A} $$ And from here, as $t$ tends to infinity, it can be seen that the solution becomes $4$

(b)For part (b) by substituting $y_{0}=2$ into the general solution, the expression becomes $$y=\frac{4}{(1+t)^{4}e^{-4t}+1} $$. I then let $y=3.99$ to get a value of $t$. Im stuck from here onwards,however, as it dont understand the rationale of the question. Espicallly for part (c). I know that as the solution becomes $4$, $T$ will tends to infinity. But what if the solution becomes $3.99$ instead. And for part (c) i dont really get what the question means when it ask to find the range of initial values for which the solution lies in the interval $3.99<y<4.01$.Could anyone explain. Thanks


$$ y = \frac{4}{1-\mathrm{e}^{-4t}(1+t)^4\left(1-\frac{4}{y_0}\right)} $$ where $y_0 = y(0)$

I think this clearer to see as you then have (divided by 4) $$ \left(1-\frac{1}{400}\right) \leq\frac{1}{1-81\mathrm{e}^{-8}\left(1-\frac{4}{y_0}\right)}\leq \left(1+\frac{1}{400}\right) $$ since $$ \left|81\mathrm{e}^{-8}\left(1-\frac{4}{y_0}\right)\right| < 1 $$ I would expand and then compare terms in expansion to resolve for $y_0$.

  • $\begingroup$ Oh thanks for the explanation, i think i got it. $\endgroup$ – ys wong Jan 22 '15 at 15:49
  • $\begingroup$ Could u please help me with this question as well. Im having problems with it. Thanks. math.stackexchange.com/questions/1118555/… $\endgroup$ – ys wong Jan 25 '15 at 7:27

Part (c)

The general solution is a family of functions. By specifying a point, like the initial solution $y(0) = y_0$, one can single out a specific solution.

Here you have the condition $3.99 < y(2) < 4.01$, which singles out a subset of solutions, where you just have to find the value of those solutions for argument $t = 0$.

Using the WA system (I am lazy here) I got the general solution

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

we see $$ y(0) = \frac{4}{e^{4c} + 1} = y_0 \iff e^{4c} = \frac{4}{y_0}-1 $$ this gives

$$ y(t) = \frac{4 e^{4 t}}{\left(\frac{4}{y_0}-1\right) (t+1)^4+e^{4 t}} $$ and $$ y(2) = y_2 = \frac{4 e^8}{81\left(\frac{4}{y_0}-1\right)+e^8} $$ solving for $y_0$ gives $$ \frac{4}{y_0}-1 = \frac{1}{81}\left(\frac{4e^8}{y_2}-e^8\right) = \frac{e^8}{81}\left(\frac{4}{y_2}-1\right) $$ or $$ y_0 = \frac{4}{\frac{e^8}{81}\left(\frac{4}{y_2}-1\right) + 1} $$ then we enter the interval boundaries as $y_2$ and get $$ 3.66 < y_0 < 4.40 $$ or similar, minus my calculation errors. :)

Part (b)

$$ y(t) = \frac{4 e^{4 t}}{\left(\frac{4}{2}-1\right) (t+1)^4+e^{4 t}} = \frac{4 e^{4 t}}{(t+1)^4+e^{4 t}} $$ so we need to solve $$ 3.99 = \frac{4 e^{4 t}}{(t+1)^4+e^{4 t}} $$ for $t$.

Using WA again, I get $t = 2.84367 = T$ as result.

Part (a)

Yes, $\lim_{t\to\infty} y(t) = 4$.

  • $\begingroup$ As $t=0$, $y=4$, which still dosent give the range of solutions. $\endgroup$ – ys wong Jan 22 '15 at 15:24
  • $\begingroup$ Oh thanks, i think i got it. $\endgroup$ – ys wong Jan 22 '15 at 15:48

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