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Question: An integrating factor of the differential equation \(x\dfrac{dy}{dx}-y={{x}^{3}};x>0\) is…… A. \(x...

An integrating factor of the differential equation xdydx−y=x3;x>0x\dfrac{dy}{dx}-y={{x}^{3}};x>0 is……
A. xx
B. −1x-\dfrac{1}{x}
C. −x-x
D. 1x\dfrac{1}{x}

Explanation

Solution

First we observe that the given differential equation is a first order linear differential equation of the form dydx+P(x)y=Q(x)\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right). The solution of a linear differential equation is given by yI.F.=∫Q(x)I.F.dxy\text{I}\text{.F}.=\int Q\left( x \right)\text{I}\text{.F}.dx where, I.F. (integrating factor) is I.F=e∫P(x)dxI.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 xdydx−y=x3;x>0x\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 xx then, the equation becomes dydx−yx=x2\dfrac{dy}{dx}-\dfrac{y}{x}={{x}^{2}}
Now, we have a differential equation, which is of the form dydx+P(x)y=Q(x)\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right).
Here, P(x)=−1xP\left( x \right)=-\dfrac{1}{x} and Q(x)=x2Q\left( x \right)={{x}^{2}}
Now, we know that the integrating factor of a linear differential equation is given by I.F=e∫P(x)dxI.F=e{}^{\int P\left( x \right)dx}
So, the integrating factor for the given equation will be I.F=e∫−1xdxI.F={{e}^{\int -\dfrac{1}{x}dx}}
First we solve ∫−1xdx\int{-\dfrac{1}{x}dx}
We know that ∫1xdx=log⁡x\int \dfrac{1}{x}dx=\log x
So, ∫−1xdx=−log⁡x\int{-\dfrac{1}{x}dx}=-\log x
⇒∫−1xdx=log⁡x−1 [As mlog⁡n=log⁡nm]\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=elog⁡x−1I.F={{e}^{\log {{x}^{-1}}}}
Now, we know that elog⁡x=xe{}^{\log x}=x so when we simplify further we get I.F=x−1=1x  \begin{aligned} & I.F={{x}^{-1}}=\dfrac{1}{x} \\\ & \\\ \end{aligned}
So the correct answer is option D.

Note: When there is the only term dydx\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.