Question

 An integrating factor of the differential equation $x\dfrac{dy}{dx}-y={{x}^{3}};x>0$ is……
A. $x$
B. $-\dfrac{1}{x}$
C. $-x$
D. $\dfrac{1}{x}$

Answer 1

Hint: First we observe that the given differential equation is a first order linear differential equation of the form $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$. The solution of a linear differential equation is given by \[y\text{I}\text{.F}.=\int Q\left( x \right)\text{I}\text{.F}.dx\] where, I.F. (integrating factor) is $I.F=e{}^{\int P\left( x \right)dx}$. We find the integrating factor by using this formula.

Complete step by Answer:

We have given that the differential equation $x\dfrac{dy}{dx}-y={{x}^{3}};x>0$, which is not in the standard form of a linear differential equation.
To convert this equation into the standard form we divide the whole equation by $x$ then, the equation becomes $\dfrac{dy}{dx}-\dfrac{y}{x}={{x}^{2}}$
Now, we have a differential equation, which is of the form $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$.
Here, $P\left( x \right)=-\dfrac{1}{x}$ and $Q\left( x \right)={{x}^{2}}$
Now, we know that the integrating factor of a linear differential equation is given by $I.F=e{}^{\int P\left( x \right)dx}$
So, the integrating factor for the given equation will be $I.F={{e}^{\int -\dfrac{1}{x}dx}}$
First we solve $\int{-\dfrac{1}{x}dx}$
We know that $\int \dfrac{1}{x}dx=\log x$
So, $\int{-\dfrac{1}{x}dx}=-\log x$
$\Rightarrow \int{-\dfrac{1}{x}dx}=\log {{x}^{-1}}\text{ }\left[ \text{As }m\log n=\log {{n}^{m}} \right]$
When we put the value the integrating factor will be $I.F={{e}^{\log {{x}^{-1}}}}$
Now, we know that $e{}^{\log x}=x$ so when we simplify further we get $\begin{align}
  & I.F={{x}^{-1}}=\dfrac{1}{x} \\
 & \\
\end{align}$
So the correct answer is option D.

Note: When there is the only term $\dfrac{dy}{dx}$ it means the equation is a first-order differential equation. The key point to solve this question is observing the given equation as a linear differential equation. The integrating factor will convert an inexact differential equation into an exact differential equation.