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I have this particular differential equation:

$$ y'+(2/3)y = 1 - 1/2t, y(0) = y_0$$

I have to find the specific value $y_0$ where the solution touches t axis, but it does not cross it.

I found the solution is

y[t] -> 1/2 (21/4 - (3 t)/2) + e^(-2 t/3) C[1] 

using mathematica, but I don't really think this function can touch zero but not crossing the t axis. Can the problem redaction be wrong??

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  • $\begingroup$ you have $1-t$ in the title and $1-1/2t$ in the body of your post; which one is right? $\endgroup$
    – abel
    May 29, 2015 at 7:57
  • $\begingroup$ I am so sorry, the correct one is 1-(1/2)t $\endgroup$
    – MSnts
    May 29, 2015 at 8:26
  • $\begingroup$ Then, fix the title, please. $\endgroup$ May 29, 2015 at 8:30

2 Answers 2

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The solution of the equation $$y'(t)+\frac 23 y(t)=1-\frac 12 t$$ with the condition $y(0)=a$ is $$y(t)=\frac{1}{8} \left((8 a-21) e^{-2 t/3}-6 t+21\right)$$ The derivative is then $$y'(t)=\frac{1}{12} \left((21-8 a) e^{-2 t/3}-9\right)$$ and this derivative cancels at $$t^*=\frac{3}{2} \log \left(\frac{7}{3}-\frac{8 a}{9}\right)$$ and $$y(t^*)=\frac{3}{2}-\frac{9}{8} \log \left(\frac{7}{3}-\frac{8 a}{9}\right)$$ Similarly, the second derivative $$y''(t^*)=-\frac{1}{2}$$ shows that the point is a maximum. So, in order for the solution to touch the $t$ axis without crossing it, what is required is $$\frac{3}{2}-\frac{9}{8} \log \left(\frac{7}{3}-\frac{8 a}{9}\right)=0$$ I am sure that you can easily take from here.

Edit

Since there is a difference between the equations in the title and in the text, let us do the same for $$y'(t)+\frac 23 y(t)=1-k t$$ with the condition $y(0)=a$. What is obtained is then $$y(t)= \frac{1}{4} \left(e^{-2 t/3} (4 a-9 k-6)+k (9-6 t)+6\right)$$ $$y'(t)=\frac{1}{6} \left(e^{-2 t/3} (-4 a+9 k+6)-9 k\right)$$ and the derivative cancel at $$t^*=\frac{3}{2} \log \left(\frac{-4 a+9 k+6}{9 k}\right)$$ for which $$y(t^*)=\frac{3}{2}-\frac{9}{4} k \log \left(\frac{-4 a+9 k+6}{9 k}\right)$$ $$y''(t^*)=-k$$ So, if $k>0$, the point corresponds to a maximum and we then need to solve for $a$ $$\frac{3}{2}-\frac{9}{4} k \log \left(\frac{-4 a+9 k+6}{9 k}\right)=0$$ $$a=\frac{1}{4} \left(6-9 \left(e^{\frac{2}{3 k}}-1\right) k\right)$$ So, the value of the maximum is given by $$y(t^*)=0$$

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  • $\begingroup$ nice to meet you on this question claude. i hope our answers match. $\endgroup$
    – abel
    May 29, 2015 at 7:35
  • $\begingroup$ Hey Claude, I am new at math, so please be a little gentle with me. I really don't know where did you get the solution where the constant "a" is involved. But I did what you told me using the equation I got first. y'=0, solving for t, then substituting this value on y. my t where y'=0 is t=(3/2)log(8c/9), so, this is the critical point of the function. So, C[1], the constant of integration involved during the solving of the diff eq is now c=(9/8)e^(10/3). Thus, y(0)=(21/8)-(9/8)e^(10/3). Do you think this is right?? $\endgroup$
    – MSnts
    May 29, 2015 at 8:38
  • $\begingroup$ $C_1$ should not exist; you must apply the condition $y(0)=a$ to get $C_1$. $\endgroup$ May 29, 2015 at 8:46
  • $\begingroup$ Oh, I see... Well, I really like your answer, (both really), but this was much help. Thank you so much. Can you help me a little with this constant "a". Just tell me some background or how is this concept called. Thank you, again. $\endgroup$
    – MSnts
    May 29, 2015 at 8:50
  • $\begingroup$ You are very welcome ! If you like it, you could accept it clicking the appropriate button. I must confess that I had a lot of fun withe problem and so I thank you. $\endgroup$ May 29, 2015 at 8:54
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suppose it touches the $t$-axis at $(t_1, 0).$ then evaluating the differential equation gives you $$t_1 = 2. $$

you can rewrite the differential equation as $$\left(ye^{2t/3}\right)' = e^{2t/3}(1-t/2) $$ integrating from $0$ to $2,$ we have $$\begin{align}-y_0 &= \int_0^2e^{2t/3}(1-t/2)\, dt \\ &= \frac32(1-t/2)e^{2t/3}\big|_0^2 + \frac34\int_0^2e^{2t/3}\, dt \\ &=-\frac32+\frac98e^{2t/3}\big|_0^2 \\ &=-\frac32+\frac98e^{4/3}-\frac98 \end{align}$$

so $$y_0 = \frac{21-9e^{4/3}}{8} $$

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  • $\begingroup$ Nice to see you too, abel ! Cheers. $\endgroup$ May 29, 2015 at 7:36
  • $\begingroup$ They don't match ! And now, the winner is .... ? $\endgroup$ May 29, 2015 at 7:47
  • $\begingroup$ @ClaudeLeibovici, how about now? $\endgroup$
    – abel
    May 29, 2015 at 7:53
  • $\begingroup$ Still not ! From my last equation $y(0)=\frac{3}{8} \left(7-3 e^{4/3}\right)$. I repeated my stuff and still get the same. Do you think I made a mistake somewhere ? I care about your opinion. $\endgroup$ May 29, 2015 at 7:58
  • $\begingroup$ @ClaudeLeibovici, i did the problem in the title with the right hand side $1-t.$ op has $1-1/2t$ in the body. that is what you did. i can go re do if that is what op wants. $\endgroup$
    – abel
    May 29, 2015 at 8:00

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