Question

If y = $\dfrac{x}{{\ln |cx|}}$ (where c is a arbitrary constant ) is the general solution of the differential equation $\dfrac{{dy}}{{dx}} = \dfrac{y}{x} + \emptyset (\dfrac{x}{y})$ then the function

$A)\,\,\,\dfrac{{{x^2}}}{{{y^2}}}$

$B)\,\,\, - \dfrac{{{x^2}}}{{{y^2}}}$

$C)\,\,\,\dfrac{{{y^2}}}{{{x^2}}}$

$D)\,\,\,\, - \dfrac{{{y^2}}}{{{x^2}}}$

Answer 1

Hint: First, differentiate the given function of y with respect to x and solve the resulted equation and convert that as an equation that is asked in the question. Also, remember that the function inside any logarithm function should always be positive. Now compare the resulted function with the function asked in question and you will get the answer. The formula that will be used in this particular question will be:

$\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$

Complete step by step answer:

Given $y=\dfrac{x}{{\ln |cx|}}$

let us assume cx is greater than 0 because in order to define the logarithmic function

therefore cx >0

then the function y is equal to $\dfrac{x}{{\ln cx}}$ where cx is assumed to be greater than 0

y= $\dfrac{x}{{\ln cx}}$ where cx > 0

by differentiating both sides of the above equation with respect to x we get

$\dfrac{{dy}}{{dx}}$ = $\dfrac{{d(\dfrac{x}{{\ln (cx)}})}}{{dx}} = \dfrac{{\ln (cx)(1) - ((x)(\dfrac{c}{{cx}}))}}{{{{\left( {\ln \left( {cx} \right)} \right)}^2}}}$  $\text{[if u and v are two functions of x , then the derivative of the quotient}$  $\dfrac{u}{v}$  $\text{is given by}$ ….

$\dfrac{{d\left( {\dfrac{u}{v}} \right)}}{{dx}} = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$ .

In words, these formulae can be remembered as :

The derivative of a quotient equals bottom times derivative of top minus top times derivative of the bottom, divided by bottom squared.]

By simplifying the above equation we get

$\dfrac{{dy}}{{dx}}$ = $\dfrac{{\ln (cx) - 1}}{{{{(\ln (cx))}^2}}}$

$\dfrac{{dy}}{{dx}}$ = $\dfrac{{\dfrac{x}{y} - 1}}{{{{\left( {\dfrac{x}{y}} \right)}^2}}} \text{            }$  {Since $y=\dfrac{x}{{\ln cx}}$ => ln(cx)= $\dfrac{x}{y}$}

Substitute $\dfrac{{dy}}{{dx}}$ in the given equation $\dfrac{{dy}}{{dx}} = \dfrac{y}{x} + \emptyset (\dfrac{x}{y})$ we get

 $\dfrac{{\dfrac{x}{y} - 1}}{{{{\left( {\dfrac{x}{y}} \right)}^2}}}$ = $\dfrac{y}{x} + \emptyset (\dfrac{x}{y})$

$\dfrac{{\dfrac{{x - y}}{y}}}{{{{\left( {\dfrac{x}{y}} \right)}^2}}} = \dfrac{y}{x} + \emptyset \left( {\dfrac{x}{y}} \right)$

$\dfrac{{\dfrac{{x - y}}{y}}}{{{{\left( {\dfrac{x}{y}} \right)}^2}}} - \dfrac{y}{x} = \emptyset (\dfrac{x}{y})$

$\emptyset (\dfrac{x}{y}) = \dfrac{{xy - {y^2}}}{{{x^2}}} - \dfrac{y}{x}$

$\emptyset (\dfrac{x}{y}) = - (\dfrac{{{y^2}}}{{{x^2}}})$

$\emptyset (\dfrac{x}{y}) = - {(\dfrac{y}{x})^2}$

Therefore in the given options option D that is $ - (\dfrac{{{y^2}}}{{{x^2}}})$ is the correct option.

Note:
Make sure to do the process of differentiation correctly. Check twice while solving the problem and also while removing the mod function check where it will be positive and where it will be negative.