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#### karush

##### Well-known member

- Jan 31, 2012

- 2,685

- Thread starter karush
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- Thread starter
- #1

- Jan 31, 2012

- 2,685

- Jan 30, 2018

- 428

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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- Jan 31, 2012

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I assume the 3 has to distributed and an exponent\)

- Jan 30, 2018

- 428

$log\left|\frac{y}{y- 4}\right|= \frac{2}{3}t^2+ C$

$\frac{y}{y- 4}= C'e^{\frac{2t^2}{3}}$

And, with $y(0)= y_0$, $\frac{y_0}{y_0- 4}=C'$.

so $\frac{y}{y- 4}= \frac{y_0}{y_0- 4}e^{\frac{2t^2}{3}}$.

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- Jan 31, 2012

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

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