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


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:


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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Are you sure your answer and the answer in the end of the book are different? – student May 19 '12 at 17:02
Yeah I couldnt simplify my answer to the one in the book. If they are the same, can you tell me how to do such simplification? – Sean87 May 19 '12 at 17:04
@Sean87: Minor comment, unrelated to your question. Early on, you have $\frac{1}{2}x^2=\frac{1}{3}e^{3t}+C$. Now is the time to find $C$, when it is naked. Not when you have buried it (solved), for then you have to unbury it. Easy here, but you will find immediate evaluation safer in general. – André Nicolas May 19 '12 at 17:45
@AndréNicolas Thanks for the tip! – Sean87 May 19 '12 at 17:49
up vote 2 down vote accepted

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 $\frac{9\cdot 146}{6}=219$

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