Particular solution of the differential equation \( (x-1)\displaystyle\frac{dy}{dx}=2xy \), when \( y(2)=1 \) is:- (consider \( \log_e x=\log x \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \log|y|-\log(x-1)^2=2x-4 \)
The variables separate:
\( \displaystyle \frac{dy}{y}=\frac{2x}{x-1}\,dx \)
Since the numerator is not of lower degree, divide 2x by (x − 1):
\( \displaystyle \frac{2x}{x-1}=\frac{2(x-1)+2}{x-1}=2+\frac{2}{x-1} \)
Integrating,
\( \log|y|=2x+2\log|x-1|+C \)
\( \log|y|-\log(x-1)^2=2x+C \)
Put x = 2, y = 1 (so log 1 = 0 and log 12 = 0):
\( 0-0=4+C \)
\( C=-4 \)
Hence \( \log|y|-\log(x-1)^2=2x-4 \).
Shortcut: Put x = 2, y = 1 into the options: both logarithms are 0, so the left side is 0. The right side is 0 only for 2x − 4, which leaves Options (2) and (3).
Differentiating Option (2),
\( \log|y|=2\log|x-1|+2x-4 \),
gives
\( \displaystyle \frac{y'}{y}=\frac{2}{x-1}+2=\frac{2x}{x-1} \),
the given equation; Option (3) gives \( -\frac{2}{x-1}+2 \) instead.