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Integrating Factor and a Linear Equation Solution | JEE

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Question
If y1(x)y_1(x) is a solution of dydx−f(x) y=0\dfrac{dy}{dx} - f(x)\,y = 0, then a solution of dydx+f(x) y=r(x)\dfrac{dy}{dx} + f(x)\,y = r(x) is:
A1y1(x)∫r(x) y1(x) dx\dfrac{1}{y_1(x)}\displaystyle\int r(x)\,y_1(x)\,dxcorrect
By1(x)∫r(x)y1(x) dxy_1(x)\displaystyle\int \dfrac{r(x)}{y_1(x)}\,dx
C∫r(x) y1(x) dx\displaystyle\int r(x)\,y_1(x)\,dx
DNone of these
Solution
Step 1: From dydx−f(x)y=0\dfrac{dy}{dx} - f(x)y = 0:
dyy=f(x) dx⇒ln⁡y=∫f(x) dx⇒y1(x)=e∫f(x) dx.\frac{dy}{y} = f(x)\,dx \Rightarrow \ln y = \int f(x)\,dx \Rightarrow y_1(x) = e^{\int f(x)\,dx}.
 Step 2: For dydx+f(x)y=r(x)\dfrac{dy}{dx} + f(x)y = r(x), the integrating factor is
IF=e∫f(x) dx=y1(x).\text{IF} = e^{\int f(x)\,dx} = y_1(x).
 Step 3: Therefore
y y1(x)=∫r(x) y1(x) dx⇒y=1y1(x)∫r(x) y1(x) dx.y\,y_1(x) = \int r(x)\,y_1(x)\,dx \Rightarrow y = \frac{1}{y_1(x)}\int r(x)\,y_1(x)\,dx.
 Correct answer: (1)
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