Question

The solution $xdy - ydx = x{y^2}dx$ is:
A. $y{x^2} + 2x + 2cy = 0$
B. ${x^2}y + 2y = 2cx$
C. $xy + x = cy$
D. ${x^2}y + 2y = c{x^3}$

Answer 1

Hint: In this question remember to simplify the given equation in terms of dx or dy and then remember to use the method of integration, using this information will help you to approach the solution of the question.

Complete step by step answer:
According to the given information we have equation; $xdy - ydx = x{y^2}dx$
Let’s simplify the given equation in terms of dx or dy
$xdy - ydx = x{y^2}dx$
$ \Rightarrow $$\dfrac{{xdy}}{{{y^2}dx}} - \dfrac{{ydx}}{{{y^2}dx}} = x$
$ \Rightarrow $$\dfrac{{xdy}}{{{y^2}dx}} - \dfrac{1}{y} = x$
$ \Rightarrow $$\dfrac{{xdy - ydx}}{{{y^2}dx}} = x$
$ \Rightarrow $$\dfrac{{ydx - xdy}}{{{y^2}dx}} = - x$
As we know that by the formula of derivative $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$
Therefore, $\dfrac{d}{{dx}}\left( {\dfrac{x}{y}} \right) = - x$
$ \Rightarrow $$d\left( {\dfrac{x}{y}} \right) = - xdx$
Now integrating both side we get
$\int {d\left( {\dfrac{x}{y}} \right)} = - \int {xdx} $
As we know that by the formula $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$ and $\int {dx} = \dfrac{{{x^{0 + 1}}}}{{0 + 1}}$
Therefore, $\dfrac{x}{y} + c = - \dfrac{{{x^2}}}{2}$
Now simplifying the above equation, we get
$\dfrac{{x + yc}}{y} = - \dfrac{{{x^2}}}{2}$
$ \Rightarrow $$2x + 2yc = - y{x^2}$
$ \Rightarrow $$2x + 2cy + y{x^2} = 0$
Therefore, solution of $xdy - ydx = x{y^2}dx$ is $y{x^2} + 2x + 2cy = 0$

So, the correct answer is “Option A”.

Note: In the above equation we simplified the given equation in terms of dx or dy which will make the equation easy so that we can apply the method of integration by applying the formula $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$ and $\int {dx} = \dfrac{{{x^{0 + 1}}}}{{0 + 1}}$ but since we had $d\left( {\dfrac{x}{y}} \right)$where we used it as $\int {d\left( {\dfrac{x}{y}} \right)} = \dfrac{{{{\left( {\dfrac{x}{y}} \right)}^{0 + 1}}}}{{0 + 1}}$ then simplified it to the simplest form, in the method of integration C represents the integration constant.