Integrating factor of the differential equation \( x(x-1)\displaystyle\frac{dy}{dx}-(x-2)y=x^3(2x-1),x\gt 1 \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{x-1}{x^2} \)
Dividing by x(x − 1) gives the linear form \( \frac{dy}{dx}+Py=Q \):
\( \displaystyle \frac{dy}{dx}-\frac{x-2}{x(x-1)}\,y=\frac{x^2(2x-1)}{x-1} \)
\( \displaystyle P=-\frac{x-2}{x(x-1)} \)
In partial fractions, \( \frac{x-2}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1} \), so \( x-2=A(x-1)+Bx \).
\( \displaystyle x=0:\ -2=-A\ \Rightarrow\ A=2 \)
\( \displaystyle x=1:\ -1=B \)
\( \displaystyle P=-\frac{2}{x}+\frac{1}{x-1} \)
\( \displaystyle \int P\,dx=-2\log x+\log(x-1)\qquad (x\gt 1) \)
\( \displaystyle =\log\frac{x-1}{x^2} \)
\( \displaystyle \text{I.F.}=e^{\int P\,dx}=\frac{x-1}{x^2} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{x+1}{x^2} \) — Incorrect. The factor from \( \frac{1}{x-1} \) is x − 1, not x + 1.
Option (2) \( \displaystyle\frac{1-x}{x^2} \) — Incorrect. \( e^{\log|x-1|}=|x-1| \), and for x > 1 this is x − 1, not 1 − x.
Option (4) \( \displaystyle\frac{x-1}{x} \) — Incorrect. The coefficient 2 in \( -\frac{2}{x} \) is lost; it gives \( x^{-2} \), not \( x^{-1} \).