Question

Particular solution of the differential equation \((x-1)\frac{dy}{dx}=2xy\), when \(y(2)=1\) is:- (consider \(\log_e x=\log x\))

• A. \(\log|y|+\log(x-1)^2=2x+4\)
• B. \(\log|y|-\log(x-1)^2=2x-4\)
• C. \(\log|y|+\log(x-1)^2=2x-4\)
• D. \(\log|y|-\log(x-1)^2=2x+4\)

Hint:

Separate variables: \(\frac{dy}{y}=\left(2+\frac{2}{x-1}\right)dx\). Then use \(y(2)=1\).

Answer 1

The Correct Option is B

Step 1: Recognise the type.
The equation can be written with all \(y\) terms on one side and all \(x\) terms on the other. It is a variable separable equation.

Step 2: Separate the variables.
\[ \frac{dy}{y}=\frac{2x}{x-1}\,dx \] Divide the numerator: \(\frac{2x}{x-1}=\frac{2(x-1)+2}{x-1}=2+\frac{2}{x-1}\).

Step 3: Integrate both sides.
\[ \log|y|=2x+2\log|x-1|+C \] Since \(2\log|x-1|=\log(x-1)^2\), this is \(\log|y|-\log(x-1)^2=2x+C\).

Step 4: Use y(2) = 1.
Put \(x=2,\ y=1\): \(\log 1-\log 1=4+C\). Both logs are 0, so \(C=-4\).

Step 5: Write the particular solution.
\[ \log|y|-\log(x-1)^2=2x-4 \]

Step 6: Check the options.
Options 1 and 3 have a plus sign between the logs, but our solution has a minus. Option 4 has \(2x+4\), which would need \(C=+4\). Only option 2 matches.

Final Answer:
The particular solution is \(\log|y|-\log(x-1)^2=2x-4\), option 2. \[ \boxed{\log|y|-\log(x-1)^2=2x-4} \]