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I have the exercise below where $x(0)=5$:

$\frac{dx}{dt}=\frac{e^{3t}}{x}$

This obviously results in (Unless I am wrong!) to:

$\int xdx = \int e^{3t}dt$ $\frac{1}{2}x^2=\frac{1}{3}e^{3t}+C$ Multiply by $2$: $x^2=\frac{2}{3}e^{3t}+2C$ $x=\sqrt{\frac{2}{3}e^{3t}+2C}$

No using $x(0)=5$ to find actual value of the constant: $25=\frac{2}{3}+2C$ $25=\frac{2+6C}{3}$

I get $C=\frac{73}{6}$ so $2C$ will be $\frac{144}{6}$

So my final answer is: $x=\sqrt{\frac{2}{3}e^{3t}+\frac{146}{6}}$

The answer in the end of the book is given as:

$x=\frac{1}{3}\sqrt{219+6e^{3t}}$

I tried this exercise quite a few time but I can't get the real answer. Could someone please tell me what I am doing wrong here?

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    @AndréNicolas Thanks for the tip! – 2012-05-19

2 Answers 2

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Start with the one from the book, for instance.

$x = \frac{1}{3}\sqrt{219 + 6e^{3t}} = \frac{\sqrt{219 + 6e^{3t}}}{3} = \sqrt{\frac{219 + 6e^{3t}}{9}},$ simply because $\sqrt{9} = 3$. Now, $\dfrac{6}{9} = \dfrac{2}{3}$, so that $x = \sqrt{\frac{219}{9} + \frac{2}{3}e^{3t}}.$ Dividing $219$ by $3$, you get $73$. So $\dfrac{219}{9} = \dfrac{73}{3} = \dfrac{146}{6}$. Thus, $x = \sqrt{\frac{219}{9} + \frac{2}{3}e^{3t}} = \sqrt{\frac{146}{6} + \frac{2}{3}e^{3t}}.$

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$\sqrt{\frac{2}{3}e^{3t}+\frac{146}{6}}=\frac{\sqrt{9}}{3}\sqrt{\frac{2}{3}e^{3t}+\frac{146}{6}}=\frac{1}{3}\sqrt{9 \left ( \frac{2}{3}e^{3t}+\frac{146}{6}\right )}=\frac{1}{3}\sqrt{6e^{3t}+219}$since $\frac{9\cdot 146}{6}=219$