Question

If $x\dfrac{{dy}}{{dx}} = y\left( {\log y - \log x + 1} \right)$, then the solution of the equation is:
A) $y\log \left( {\dfrac{x}{y}} \right) = cx$
B) $x\log \left( {\dfrac{y}{x}} \right) = cy$
C) $\log \left( {\dfrac{y}{x}} \right) = cx$
D) $\log \left( {\dfrac{x}{y}} \right) = cy$

Answer 1

Hint : In the given question, we are given a differential equation. So, we have to solve the given differential equation using methods of integration. Then, we will find the value of the expression provided to us in the question itself. We first convert the given differential equation into a form that is easier to solve compared to the one given in the problem itself with the use of substitution.

Complete step-by-step answer :
The given question requires us to solve the differential equation $x\dfrac{{dy}}{{dx}} = y\left( {\log y - \log x + 1} \right)$ using methods of integration.
So, we will first use the law of logarithm $\log a - \log b = \log \dfrac{a}{b}$ to simplify the expression on the right side of the equation.
So, we have, $x\dfrac{{dy}}{{dx}} = y\left( {\log y - \log x + 1} \right)$
$\Rightarrow x\dfrac{{dy}}{{dx}} = y\left( {\log \dfrac{y}{x} + 1} \right)$
Shifting all the terms except the differential $\dfrac{{dy}}{{dx}}$ to the right side of equation, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{x}\left( {\log \dfrac{y}{x} + 1} \right) - - - - \left( 1 \right)$
Now, we substitute the expression $\dfrac{y}{x}$ as t. So, we get, $y = tx$. Differentiating both sides with respect to t, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {tx} \right)$
Now, we use the product rule of differentiation $\dfrac{d}{{dx}}\left( {f(x) \times g(x)} \right) = f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right) + g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right)$ . So, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = t\dfrac{d}{{dx}}\left( x \right) + x\dfrac{d}{{dx}}\left( t \right)$
Now, we know the power rule of differentiation as $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$. So, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = t + x\dfrac{{dt}}{{dx}}$
Now, substituting this value in the differential equation $\left( 1 \right)$ , we get,
$\Rightarrow t + x\dfrac{{dt}}{{dx}} = \dfrac{y}{x}\left( {\log \dfrac{y}{x} + 1} \right)$
Also, substituting $\dfrac{y}{x} = t$.
$\Rightarrow t + x\dfrac{{dt}}{{dx}} = t\left( {\log t + 1} \right)$
Opening brackets, we get,
$\Rightarrow t + x\dfrac{{dt}}{{dx}} = t\log t + t$
Cancelling the like terms with same sign on both sides of equation,
$\Rightarrow x\dfrac{{dt}}{{dx}} = t\log t$
Now, we separate the variables from one another. So, we get,
$\Rightarrow \dfrac{{dt}}{{t\log t}} = \dfrac{{dx}}{x}$
Now, we can integrate both sides directly as the variables in the differential equation have been separated. So, we get,
$\Rightarrow \int {\dfrac{{dt}}{{t\log t}}} = \int {\dfrac{{dx}}{x}}$
We know that the integral of $\dfrac{1}{x}$ with respect to x is $\ln x$ .
$\Rightarrow \int {\dfrac{{dt}}{{t\log t}}} = \ln x + \ln c$, where c is any arbitrary constant.
Also, substituting the value of $\log t$ as r, we get, $r = \log t$ . Differentiation both sides with respect to t, we get,
$\dfrac{{dr}}{{dt}} = \dfrac{1}{t}$
So, substituting the value of $\dfrac{{dt}}{t}$ as $dr$ , we get,
$\Rightarrow \int {\dfrac{{dr}}{r}} = \ln x + \ln c$
Simplifying the integral,
$\Rightarrow \ln r = \ln x + \ln c$
Now, we know the logarithmic property as $\log a + \log b = \log ab$ . So, we get,
$\Rightarrow \ln r = \ln \left( {cx} \right)$
Taking the exponential function on both sides, the logarithmic function is removed as both the functions are inverse of each other. Hence, we get,
$\Rightarrow r = \left( {cx} \right)$
Now, substituting back the value of r, we get,
$\Rightarrow \left( {\log t} \right) = \left( {cx} \right)$
Also, substituting value of t,
$\Rightarrow \log \dfrac{y}{x} = cx$
Hence, the solution to the differential equation $x\dfrac{{dy}}{{dx}} = y\left( {\log y - \log x + 1} \right)$ is $\log \dfrac{y}{x} = cx$. Hence, option (C) is the correct answer.
So, the correct answer is “Option C”.

Note : The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable with one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the constant. The exact value of the arbitrary constant can be calculated only if we are given a point lying on the function.