Question

The solution of the differential equation $x \frac{\text{d}^2 y}{\text{d}x^2} = 1$ at $x = y = 1$ with $\frac{\text{d}y}{\text{d}x} = 0$ at $x = 1$, is

• A. $y = x \log x + x + 2$
• B. $y = x \log x - x + 2$
• C. $x = x \log x + 2$
• D. $x \log x - x = y$

Hint:

$\int \ln x\,dx = x\ln x - x$

Answer 1

The Correct Option is B

Step 1: Given equation. \[ x \frac{d^2y}{dx^2} = 1 \Rightarrow \frac{d^2y}{dx^2} = \frac{1}{x} \]
Step 2: Integrate once. \[ \frac{dy}{dx} = \int \frac{1}{x} dx = \ln x + C_1 \] Using $\frac{dy}{dx} = 0$ at $x=1$: \[ 0 = \ln 1 + C_1 \Rightarrow C_1 = 0 \]
Step 3: Integrate again. \[ y = \int \ln x\,dx = x\ln x - x + C_2 \] Using $y(1)=1$: \[ 1 = 0 -1 + C_2 \Rightarrow C_2 = 2 \]
Step 4: Conclusion. \[ y = x\ln x - x + 2 \]