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I have the equation below:

$$(t^2 + 1)dx=(x+4)dt$$

Where $x(0) = 3$

I am trying to use separation of variables, and I end up here:

$$\ln(x+4)=\arctan(t)+C$$

Trying to simplify it more:

$$x=-4+\ln(\arctan(t)+C)$$

Is this correct? I think I should use $x(0) = 3$ to find value of the constant, how can I do that?

Thanks

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  • $\begingroup$ From line 2 to line 3 went something wrong. I guess you wanted to exponentiate both sides? $\endgroup$ – Fabian May 14 '12 at 12:10
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    $\begingroup$ @Sean87 I like the writing on the cup (Feel the same way about changing the world) $\endgroup$ – Kirthi Raman May 14 '12 at 14:01
  • $\begingroup$ Good :P but the source is closed ;) $\endgroup$ – Sean87 May 14 '12 at 15:16
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You are right when you got $\ln(x+4)=\arctan(t)+C$. Starting from here, since $e^{\ln x}=x$, we have $$x+4=e^{\ln(x+4)}=e^{\arctan(t)+C}=e^C\cdot e^{\arctan(t)}=C_1e^{\arctan(t)}$$ where $C_1=e^C$. By the initial condition $x(0)=3$, we have $$7=C_1e^{\arctan(0)}=C_1.$$ Therefore, $$x=7e^{\arctan(t)}-4$$ is the solution of the initial value problem.

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    $\begingroup$ It should be $7 e^{arctan(t)}$ because $x(0)=3$ not $4$. You still have +1 $\endgroup$ – Kirthi Raman May 14 '12 at 12:36
  • $\begingroup$ @Artin: Thanks. I edited it. $\endgroup$ – Paul May 14 '12 at 12:45
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$$ \frac{\mathrm{d}x}{x+4} = \frac{\mathrm{d}t}{t^2+1}$$

Integrating both sides

$$ \ln (x+4) = arctan(t) + C \tag{1}$$

$x(0) = 3$ implies

$$ \ln(7) = C$$

Rewriting $(1)$

$$ \begin{align*} \ln (x+4) &= arctan(t) + \ln(7) \\ \ln (x+4) - \ln (7) &= arctan(t)\\ \ln \frac{x+4}{7} &= arctan(t)\\ \frac{x+4}{7} = e^{arctan(t)}\\ \Rightarrow x = -4 + 7 e^{arctan(t)} \end{align*} $$

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  • $\begingroup$ For full credit you should probably say why the other possibility $$\ln(-x-4) = \arctan(t)+C$$ is not the one to use. $\endgroup$ – GEdgar May 14 '12 at 13:28
  • $\begingroup$ Yes, of course (Thanks) $\endgroup$ – Kirthi Raman May 14 '12 at 14:00
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$x(0)=3$ means when $t = 0$ , $x(t) = x = 3$ here $x$ is function in $t$

Here you have to find value of $C$

$\log(x + 4) = \arctan(t) + C$

$\log(3 + 4) = \arctan(0) + C$; $C = \log(7)$

Therefore, the solution is $\log(x + 4) = \arctan(t) + \log(7)$

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  • $\begingroup$ Looks better in TeX, doesn't it? Have a look at how I did it, and then you will be able to do it yourself. $\endgroup$ – Gerry Myerson May 14 '12 at 12:32
  • $\begingroup$ To see the code behind the Tex code click on edit link just below the answer. $\endgroup$ – Tomarinator May 14 '12 at 12:36
  • $\begingroup$ @Prasad, also mention that $\log()$ here is to the base $e$ otherwise one can use $\ln()$ $\endgroup$ – Kirthi Raman May 14 '12 at 12:39
  • $\begingroup$ Another way to see the $\TeX$ code is to highlight the expression, right-click (or Mac equivalent), select "Show Math As" and then "TeX Commands". You might copy what you see straight to your edit box and then modify it to suit your needs (after enclosing in $'s, since the display omits those). $\endgroup$ – David Lewis May 14 '12 at 12:47

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