The differential equation corresponding to primitive $y=e^{cx}$ or the elimination of the arbitrary constant m from the equation $y=e^{mx}$ gives the differential equation
A. $\dfrac{\text{d}y}{\text{d}x}=\lgroup\dfrac{y}{x}\rgroup\log~x$
B. $\dfrac{\text{d}y}{\text{d}x}=\lgroup\dfrac{x}{y}\rgroup\log~y$
C. $\dfrac{\text{d}y}{\text{d}x}=\lgroup\dfrac{y}{x}\rgroup\log~y$
D. $\dfrac{\text{d}y}{\text{d}x}=\lgroup\dfrac{x}{y}\rgroup\log~x$

Hint: First take logarithm on the both sides of the equation given in the question. Further, simplify it to get the value of the arbitrary constant which needs to be eliminated. Now, take the differentiation of the given equation and do the needful substitution which will then eliminate the arbitrary constant. This will give the required differential equation.

Formula Used: $\log~x^{n}=n~\log~x$
$\dfrac{\text{d}\lgroup~e^{x}\rgroup}{\text{d}x}=e^{x}$

Complete step by step solution: According to the query, the given equation is $y=e^{mx}$......(i)
We know,
$\log~x^{n}=n~\log~x$......(ii)
By taking the log (considering the base to be e) on both sides of the above equation, we can write
$\log~y=mx~\log~e$ [Using equation (ii)]
Now,
$\log~y=mx$ [Since, log e=1 if the base taken is e]
This implies
$m=\dfrac{\log~y}{x}$........(iii)
As it is known that $\dfrac{\text{d}\lgroup~e^{x}\rgroup}{\text{d}x}=e^{x}$......(iv)
Now, let's differentiate equation (i) with respect to x,
$\dfrac{\text{d}y}{\text{d}x}=me^{mx}$ [using (iv)]
Now, replace the value of m using equation (iii)
$\dfrac{\text{d}y}{\text{d}x}=\dfrac{\log~y}{x}e^{mx}$
Using (i) the above equation can be written as,
$\dfrac{\text{d}y}{\text{d}x}=\dfrac{\log~y}{x}y$
This implies,
 $\dfrac{\text{d}y}{\text{d}x}=\lgroup\dfrac{y}{x}\rgroup\log~y$

Option ‘C’ is correct

Note: Whenever there is two arbitrary constants in the mentioned equation, ensure to differentiate the equation two times. The reason behind doing this is to eliminate the arbitrary constant.