Question

I have " \frac{dx}{dt} = \frac{5}{x} " alt=" \frac{dx}{dt} = \frac{5}{x} " align="absmiddle" class="latex-formula"> and $x(0)=7$ and I'm supposed to find x(t)
I did it as follows:
$x dx=5dt$
$\int\ {x} \, dx = \int\ {5} \, dt$
$\frac{x^2}{2} = 5t +c$
$\frac{7^2}{2} = 5(0) +c$
And from this i obtain $c= \frac{49}{2}$ but apparently this is not correct.

Answer 1

$\bf \cfrac{dx}{dt}=\cfrac{5}{t}\quad ,\quad x(0)=7\to \begin{cases} f(t)=0\\ t=7 \end{cases}\\\\ -----------------------------\\\\ \displaystyle \int \cfrac{5}{t}\cdot dt\implies 5\int \cfrac{1}{t}\cdot dt\implies 5ln|t|+C= f(t) \\\\\\ \textit{now, we know when f(t)=0, t=7}\implies 5ln|7|+C=0 \\\\\\ C=-5ln(7) \\\\\\ thus\implies \boxed{5ln|t|-5ln(7)=f(t)=x}$

notice |7| 7 is positive, so we can simply remove the bars and use 7 by itself