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: Predict the form of the solution.
The equation is separable, so its general solution has the form $y=C(x-1)^2e^{2x}$. We will confirm this and then fit the condition.

Step 2: Derive that form.
From $\frac{dy}{y}=\left(2+\frac{2}{x-1}\right)dx$, integrating gives $\ln|y|=2x+2\ln|x-1|+k$. Taking exponentials gives $y=C(x-1)^2e^{2x}$.

Step 3: Find C.
Use $y(2)=1$: $1=C\cdot 1\cdot e^{4}$, so $C=e^{-4}$.

Step 4: Take logs again.
$\ln|y|=-4+2x+\ln(x-1)^2$. Rearranging gives $\ln|y|-\ln(x-1)^2=2x-4$.

Step 5: Sanity check.
At $x=2$: left side is $\ln 1-\ln 1=0$ and right side is $4-4=0$. This holds. For option 4 the right side would be $8$, which fails.

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