# [SOLVED]b.2.2.27 IVP t behavior

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#### Country Boy

##### Well-known member
MHB Math Helper
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.

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#### karush

##### Well-known member
for some reason your latex didn't render with [tex] I thot it was supposed to be $$\displaystyle {/math] or  I assume the 3 has to distributed and an exponent$$

#### Country Boy

##### Well-known member
MHB Math Helper
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

#### Country Boy

##### Well-known member
MHB Math Helper
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 $-0.1143e^{-\frac{2t^2}{3}}= -199$.

• karush