Author Topic: Q1: TUT0401  (Read 1456 times)

Victor Ivrii

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Q1: TUT0401
« on: September 28, 2018, 03:34:12 PM »
Consider the initial value problem
\begin{equation*}
y' + \frac{2}{3} y = 1 - \frac{1}{2} t,\qquad y(0) = y_0.
\end{equation*}
Find the value of $y_0$ for which the solution touches, but does not cross, the $t$-axis.

Ruoyu Zhang

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Re: Q1: TUT0401
« Reply #1 on: September 28, 2018, 06:09:01 PM »
TUT0401

Victor Ivrii

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Re: Q1: TUT0401
« Reply #2 on: September 29, 2018, 03:19:47 PM »
Too large picture, difficult to read

Wei Cui

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Re: Q1: TUT0401
« Reply #3 on: September 29, 2018, 09:16:33 PM »
tut0401   $y^{'} + \frac{2}{3}y=1-\frac{1}{2}t$, $y(0) = y_0$
   
$p(t) = \frac{2}{3}$, $g(t) = 1-\frac{1}{2}t$, then $u = e^{\int \frac{2}{3}dt}$,

multiply both sides with $u$, then we get:

$e^{\frac{2}{3}t}y^{'} + \frac{2}{3}e^{\frac{2}{3}t}y = e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t}$

$(e^{\frac{2}{3}t}y)^{'}=e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t}$

$d(e^{\frac{2}{3}t}y) = (e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t})dt$

$e^{\frac{2}{3}t}y=\int (e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t})dt$

$e^{\frac{2}{3}t}y=\frac{3}{2}e^{\frac{2}{3}t}-\frac{3}{4}te^{\frac{2}{3}t}+\frac{9}{8}e^{\frac{2}{3}t}+C$    (integrate by parts)

$y= \frac{3}{2}-\frac{3}{4}t+\frac{9}{8}+Ce^{-\frac{2}{3}t}$

$y=\frac{21}{8}-\frac{3}{4}t+Ce^{-\frac{2}{3}t}$. Consider $y(0) = y_0 \implies y_0=\frac{21}{8}-0+C \implies C=y_0-\frac{21}{8}$

Therefore, $y=\frac{21}{8}-\frac{3}{4}t+(y_0-\frac{21}{8})e^{-\frac{2}{3}t}$

If $y(t)$ touches, but does not cross the $t$-axis at some point $t_0$ s.t. $y(t_0) = 0$ and $y^{'}(t_0) = 0$

Thus, $y^{'} +\frac{2}{3}y = 1- \frac{1}{2}t$ when $t=t_0$

$0+0 = 1-\frac{1}{2}t_0$

$\frac{1}{2}t_0 = 1 \implies t_0 = 2$

substitute $t_0 = 2$, then $y(2) = 0 \implies 0 = \frac{21}{8} - \frac{3}{4}\times 2 + (y_0-\frac{21}{8})e^{-\frac{4}{3}}$

Therefore, we get $y_0 = \frac{21}{8}e^{-\frac{4}{3}}-\frac{9}{8}$