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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

  • 0
    Good :P but the source is closed ;)2012-05-14

3 Answers 3

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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.

  • 0
    @Artin: Thanks. I edited it.2012-05-14
5

$ \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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    Yes, of course (Thanks)2012-05-14
4

$x(0)=3$ 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 $\log(x + 4) = \arctan(t) + \log(7)$

  • 0
    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).2012-05-14