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I have the following ODE Where given is $x(0)=1$:

$$(t+3)dx=4x^2dt$$

After separation of variables I got this:

$$\frac{-1}{x} = 4\ln(t+3)+C$$

I think this simplifies more as:

$$x=\frac{-1}{\ln((t+3)^4)+C}$$

Please tell me if this is correct, I also have problem finding C in this case, MapleTA does not accept my answer which happens to be:

$$x=\frac{-1}{\ln((t+3)^4)+(-0.52)}$$

Update:

I found that the C should e expresed also in $\ln$ (logarithmic) term, not a number. Any ideas?

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To find the value of $C$, plug in the initial conditions: plugging in $t=0$ and setting $x=1$ should allow you to obtain the value for $C$. –  Arturo Magidin May 14 '12 at 19:30

1 Answer 1

up vote 1 down vote accepted

The general solution seems correct, but your computation of $C$ does not.

You know that $x(0)=1$. That means that when $t=0$, you should get $x=1$. Plugging in $t=0$ you get $$x(0) = \frac{-1}{4\ln(0+3) + C} = \frac{-1}{4\ln(3)+C}.$$ Since $x(0)=1$, that means $$\begin{align*} 1 &= \frac{-1}{4\ln(3)+C}\\ 4\ln(3)+C &= -1\\ C &= -1-4\ln(3)\approx -1-4(1.0986) = -5.3944 \end{align*}$$ So I'm not sure how you get that $C\approx -0.52$.

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Thanks, I contacted a fellow classmate, What I have to achive here is not a real number, it should be expressed in terms of logarithms. Any idea how to do that? –  Sean87 May 14 '12 at 22:03
    
Aaaaa I see you already did that :P I did too but with a little bit of mistake! Thanks! –  Sean87 May 14 '12 at 22:56

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