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 An…

Question

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.

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 \boxedt`

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