Question

What is the solution of the differential equation $x\dfrac{{dy}}{{dx}} = y(\log y - \log x + 1)$?

Answer 1

Hint: The differential equation is the equation that combines the function and its derivatives. Here we need to find the solution of the given differential equation. The solution of a differential equation is nothing but an expression for the dependent variable in terms of independent variables that satisfy the equation; the unknown function is the dependent variable and the variables on which it depends are independent variables.
Formula:
$\log a - \log b = \log \dfrac{a}{b}$

Complete Step by step solution:
Given,
The equation to find the solution of the equation $x\dfrac{{dy}}{{dx}} = y(\log y - \log x + 1)$.
$x$ can be divided from the right and left sides of the equation.
$\dfrac{{dy}}{{dx}} = \dfrac{y}{x}(\log y - \log x + 1)$
As we know $\log a - \log b = \log \dfrac{a}{b}$
As we substitute this formula in the equation.
$\dfrac{{dy}}{{dx}} = \dfrac{y}{x}(\log \dfrac{y}{x} + 1).........(1)$
Let substitute $y = vx$
Differentiate this above equation with respect to $x$.
The formula for differentiation ${(uv)'} = {u'}v + u{v'}$
Differentiating $y = vx$equation by ${(uv)'} = {u'}v + u{v'}$
$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}...........(2)$
By equating equation $(1)$ and $(2)$, we get
$\dfrac{y}{x}(\log \dfrac{y}{x} + 1) = v + x\dfrac{{dv}}{{dx}}$
Substitute $v = \dfrac{y}{x}$ in the above equation
$v(\log v + 1) = v + x\dfrac{{dv}}{{dx}}$
Multiply $v$inside the left side of the equation,
$v\log v + v = v + x\dfrac{{dv}}{{dx}}$
Take $v$terms on one side and $x$terms on the other side, we get
$v\log v + v - v = x\dfrac{{dv}}{{dx}}$
By subtracting the above equation, we get
$v\log v = x\dfrac{{dv}}{{dx}}$
We should take $v$terms on one side and $x$terms on the other side, we get
$\dfrac{{dv}}{{v\log v}} = \dfrac{{dx}}{x}$
By integrating both sides, we get,
$\int {\dfrac{{dv}}{{v\log v}}} = \int {\dfrac{{dx}}{x}} ......(3)$
Let us assume that${I_1} = \int {\dfrac{{dv}}{{v\log v}}} $
 ${I_2} = \int {\dfrac{{dx}}{x}} $
Let us integrate the ${I_1}$term,
${I_1} = \int {\dfrac{{dv}}{{v\log v}}} $
Let us assume $u = \log v$
Differentiating u with respect to v we get,
$\dfrac{{du}}{{dv}} = \dfrac{1}{v}$
The terms in the denominator of the left side equation go to the numerator of the right side equation
$du = \dfrac{1}{v}dv$
We can substitute $du = \dfrac{1}{v}dv$in ${I_1} = \int {\dfrac{{dv}}{{v\log v}}} $
${I_1} = \int {\dfrac{{du}}{u}} $
As we integrate the above equation, we get
${I_1} = \log u$
Substitute $u = \log v$in ${I_1} = \log u$
${I_1} = \log (\log v) + c$
Integrating ${I_2} = \int {\dfrac{{dx}}{x}} $, we get
${I_2} = \log x + c$
Substitute ${I_1} = \log (\log v)$and ${I_2} = \log x$in $(3)$
$\log (\log v) = \log x........(4)$
Substitute $v = \dfrac{y}{x}$in the equation $(4)$, we get
$\log (\log \dfrac{y}{x}) = \log x$
Move the terms on the right side to the left side of the equation,
$\log (\log \dfrac{y}{x}) - \log x = 0$
Substitute $\log a - \log b = \log \dfrac{a}{b}$in $\log (\log \dfrac{y}{x}) - \log x = 0$
$\log (\log y - \log x) - \log x = 0$
Canceling $\log $on both sides,
$(\log y - \log x) = x$
Substitute $\log a - \log b = \log \dfrac{a}{b}$
$(\log \dfrac{y}{x}) = x$
Converting $\log $to $e$, we had
$\dfrac{y}{x} = {e^x}$
Multiply the terms in the denominator of the left side equation goes to the numerator of right side equation,
$y = x{e^x} + c$
The solution of the differential equation $x\dfrac{{dy}}{{dx}} = y(\log y - \log x + 1)$is $y = x{e^x} + c$.

Note:
The method for differential equation should correctly be selected. First, check whether the equation is the homogeneous or non-homogeneous solution. The integration should be known well. Properly do differentiation and integration. The formula for integration and differentiation should be known correctly. The solution for the differential equation should not be in derivatives.