MHB What is the solution for t when $-0.1143e^{-\frac{2t^2}{3}}= -199$?

[karush]

$y'=ty\dfrac{4-y}{3};\quad y{0}=y_0$
separate

$\dfrac{3}{y(4-y)}\ dy=t\ dt$
integrate
$3\left(\frac{1}{4}\ln \left|y\right|-\frac{1}{4}\ln \left|y-4\right|\right)=\dfrac{t^2}{2}+c$

hopefully so far...

actually it kinda foggy what they are eventually asking for

also why is t in different cases

 

[HOI]

Yes, that is correct so far! Now, you might use the facts that $rlog(x)= log(x^r)$ and $log(a)- log(b)= log(a/b)$. And you can get rid of the logarithm by taking the exponential on both sides.

 

[karush]

for some reason your latex didn't render with I thot it was supposed to be [math]{/math] or $$<br /> <br /> I assume the 3 has to distributed and an exponent[/math]

 

[HOI]

Write it as $\frac{3}{4}log\left|\frac{y}{y- 4}\right|= t^2/2+ c$
$log\left|\frac{y}{y- 4}\right|= \frac{2}{3}t^2+ C$
$\frac{y}{y- 4}= C'e^{\frac{2t^2}{3}}$

And, with $y(0)= y_0$, $\frac{y_0}{y_0- 4}=C'$.
so $\frac{y}{y- 4}= \frac{y_0}{y_0- 4}e^{\frac{2t^2}{3}}$.

 

[karush]

ok here is the book answer to (b)
assume various inputs were tried to get to 3.98

 

[HOI]

Part (b) told you that $x_0= 0.5$ so $\frac{y_0}{y_0-4}= \frac{0.5}{0.5- 4}= \frac{0.4}{-3.5}= -.01143$.

$\frac{y}{y- 4}= -0.1143e^{\frac{2t^2}{3}}$.

We want y to be 3.98 so $\frac{y}{y- 4}= \frac{3.98}{3.98- 4}= -\frac{3.98}{0.02}= -199
$.

Solve $-0.1143e^{-\frac{2t^2}{3}}= -199$.