Question

he integrating factor of the differential equation $\left( y\log y \right)dx=\left( \log y-x \right)dy$ is
A. $\dfrac{1}{\log y}$
B. $\log (\log y)$
C. $1+\log y$
D. $\dfrac{1}{\log (\log y)}$
E. $\log y$

Answer 1

Hint: Here, in the given question, we need to find the integrating factor of the differential equation $\left( y\log y \right)dx=\left( \log y-x \right)dy$. To find the integrating factor of the given equation $\left( y\log y \right)dx=\left( \log y-x \right)dy$, first we have to observe the type of given differential equation, which is linear differential equation $\dfrac{dx}{dy}+Rx=S$. Then we will find its integrating factor as ${{e}^{\int{R\,dy}}}$ to get our required answer.

Formula Used:
Integrating Factor = ${{e}^{\int{R\,dy}}}$
Linear Differential Equation = $\dfrac{dx}{dy}+Rx=S$

Complete step by step Solution:
We have, $\left( y\log y \right)dx=\left( \log y-x \right)dy$
$\Rightarrow \dfrac{dx}{dy}=\dfrac{\left( \log y-x \right)}{y\log y}$
This above written equation can also be written as
$\Rightarrow \dfrac{dx}{dy}=\dfrac{\log y}{y\log y}-\dfrac{x}{y\log y}$
On canceling the common factors, we get
$\Rightarrow \dfrac{dx}{dy}=\dfrac{1}{y}-\dfrac{x}{y\log y}$
$\Rightarrow \dfrac{dx}{dy}+\dfrac{x}{y\log y}=\dfrac{1}{y}\,\,..........\left( i \right)$
The above written equation is in the form of linear differential equation: $\dfrac{dx}{dy}+Rx=S$ where $R$ and $S$ are functions of $y$ or constants. Integrating factor for equation $\dfrac{dx}{dy}+Rx=S$ is given by ${{e}^{\int{R\,dy}}}$.
For equation, $\dfrac{dx}{dy}+\dfrac{1}{y\log y}\times x=\dfrac{1}{y}$, $R=\dfrac{1}{y\log y}$ and $S=\dfrac{1}{y}$
$\Rightarrow I.F.={{e}^{\int{R\,dy}}}$
Now, substitute the value of $R$
$\Rightarrow I.F.={{e}^{\int{\dfrac{1}{y\log y}\,dy}}}\,\,......\left( ii \right)$
Let $\log y=t$.
Now, differentiate the above written equation w.r.t. $t$. As we know $\dfrac{d}{dx}\left( {{\log }_{e}}x \right)=\dfrac{1}{x}$. Therefore, we get
$\Rightarrow \dfrac{1}{y}=\dfrac{dt}{dy}$
$\Rightarrow \dfrac{1}{y}dy=dt$
Now, substitute the value of $\dfrac{1}{y}dy$ and $\log y$ in equation $\left( ii \right)$.
$\Rightarrow I.F.={{e}^{\int{\dfrac{1}{t}\,dy}}}$
As we know $\int{\dfrac{1}{x}}dx={{\log }_{e}}\left| x \right|+C$. Therefore, we get
$\Rightarrow I.F.={{e}^{{{\log }_{e}}\left| t \right|+C}}$
On substituting value of $t$, we get
$\Rightarrow I.F.={{e}^{{{\log }_{e}}\left| \log y \right|+C}}$
$\Rightarrow I.F.=\log y$
Hence, the integrating factor of the given differential equation $\left( y\log y \right)dx = \left( \log y-x \right)dy$ is $\log y$.

Therefore, the correct answer is option 5.

Note: While solving these types of questions, firstly re-arrange the differential equation in such a way that you can get a differential equation in some known form such as variable separable, homogeneous or linear differential equation. While finding an integrating factor or a particular solution, find the value of constant very accurately as a particular solution is a unique solution.