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(A) $x = C{e^{my}}$ (B) $x = C{e^{ - my}}$ (C) $x = my + C$ (D) $x = C$

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

Complete step-by-step answer:

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

If we separate $x$ and $y$ terms, we’ll get:

$

\Rightarrow \dfrac{{dx}}{{dy}} + mx = 0, \\

\Rightarrow \dfrac{{dx}}{{dy}} = - mx, \\

\Rightarrow \dfrac{{dx}}{x} = - mdy \\

$

Integrating both sides, we’ll get:

$ \Rightarrow \int {\dfrac{{dx}}{x}} = - m\int {dy} ,$

We know that, $\int {\dfrac{{dx}}{x}} = \ln x$. Using this we’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, we’ll get the same result.

Complete step-by-step answer:

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

If we separate $x$ and $y$ terms, we’ll get:

$

\Rightarrow \dfrac{{dx}}{{dy}} + mx = 0, \\

\Rightarrow \dfrac{{dx}}{{dy}} = - mx, \\

\Rightarrow \dfrac{{dx}}{x} = - mdy \\

$

Integrating both sides, we’ll get:

$ \Rightarrow \int {\dfrac{{dx}}{x}} = - m\int {dy} ,$

We know that, $\int {\dfrac{{dx}}{x}} = \ln x$. Using this we’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, we’ll get the same result.