The differential equation corresponding to th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation $y = e^{mx}$.Taking the natural logarithm on both sides,we get $\log y = mx$.From this,we can express the arbitrary constant as

Vedclass · Accepted Answer

$\frac{dy}{dx} = \left( \frac{y}{x} \right) \log y$

Vedclass · Answer

(C) Given the equation $y = e^{mx}$.Taking the natural logarithm on both sides,we get $\log y = mx$.From this,we can express the arbitrary constant as $m = \frac{\log y}{x}$.Now,differentiate $y = e^{mx}$ with respect to $x$:$\frac{dy}{dx} = m e^{mx}$.Since $e^{mx} = y$,we substitute this into the derivative:$\frac{dy}{dx} = m \cdot y$.Substitute the value of $m = \frac{\log y}{x}$ into the equation:$\frac{dy}{dx} = \left( \frac{\log y}{x} \right) \cdot y = \left( \frac{y}{x} \right) \log y$.Thus,the correct option is $C$.

The differential equation corresponding to the primitive $y = e^{mx}$ is obtained by eliminating the arbitrary constant $m$. What is the resulting differential equation?

Similar Questions

The differential equation of the family of curves $r^2 = a^2 \cos 2\theta$,where '$a$' is an arbitrary constant,is:

If $m$ and $n$ are the order and degree of the differential equation of the family of parabolas with focus at the origin and $X$-axis as its axis,then $m n-m+n=$

If $a$ and $b$ are arbitrary constants,then the differential equation having $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as its general solution is

The differential equation of the family of all circles of radius $a$ is

If the differential equation obtained by eliminating $A$ and $B$ from $y = (\sin^{-1} x)^2 + A \cos^{-1} x + B$ is $(a - x^2) y'' - x y' = b$,then $\frac{b + a}{b - a} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module