Question

# Solution of $\dfrac{{dx}}{{dy}} + mx = 0,{\text{ where }}m < 0$ is:(A) $x = C{e^{my}}$ (B) $x = C{e^{ - my}}$ (C) $x = my + C$ (D) $x = C$

Hint: Separate the terms containing $x$ and $y$ on either side and then integrate both sides.

According to the question, the given differential equation is $\dfrac{{dx}}{{dy}} + mx = 0$.

If we separate $x$ and $y$ terms, weâ€™ll get:
$\Rightarrow \dfrac{{dx}}{{dy}} + mx = 0, \\ \Rightarrow \dfrac{{dx}}{{dy}} = - mx, \\ \Rightarrow \dfrac{{dx}}{x} = - mdy \\$
Integrating both sides, weâ€™ll get:
$\Rightarrow \int {\dfrac{{dx}}{x}} = - m\int {dy} ,$
We know that, $\int {\dfrac{{dx}}{x}} = \ln x$. Using this weâ€™ll get:
$\Rightarrow \ln x = - my + c,{\text{ where }}c{\text{ is the constant of integration}} \\ \Rightarrow x = {e^{ - my + c}}, \\ \Rightarrow x = {e^c} \times {e^{ - my}}, \\ \Rightarrow x = C{e^{ - my}}{\text{ [}}\therefore {e^c}{\text{ = C (another constant) ]}} \\$
Thus, the solution of the given differential equation is $x = C{e^{ - my}}$. (B) is the correct option.

Note: The above differential equation is a first order differential equation. We can also solve it by calculating integrating factors. Suppose we have a first order differential equation:

$\Rightarrow \dfrac{{dx}}{{dy}} + Px = Q,$
We calculate the integrating factor as:

$\Rightarrow I = {e^{\int {Pdy} }}$.

The solution of the differential equation is:
$\Rightarrow Ix = \int {IQdy}$
If we solve by this method, weâ€™ll get the same result.