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5 votes
I have
(dx)/(dt) = (5)/(x) 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

(x^2)/(2) = 5t +c

(7^2)/(2) = 5(0) +c
And from this i obtain
c= (49)/(2) but apparently this is not correct.

User Marc Eaddy
by
6.4k points

1 Answer

7 votes

\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 \boxedt

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

User Phil Ninan
by
5.0k points