Step 1: Understanding the Concept:
We are given a second-order linear differential equation that can be solved by direct successive integration. We will integrate once to find the first derivative $\frac{dy}{dx}$, use the first initial condition to find the constant of integration, and then integrate again to find $y(x)$, using the second initial condition to find the final constant.
Step 2: Key Formula or Approach:
1. Rewrite the equation: $\frac{d^2y}{dx^2} = \frac{1}{x}$.
2. Integrate with respect to $x$: $\int \frac{d^2y}{dx^2} dx = \int \frac{1}{x} dx \implies \frac{dy}{dx} = \log|x| + c_1$.
3. Integrate again: $\int \frac{dy}{dx} dx = \int (\log x + c_1) dx$. (Use integration by parts for $\log x$).
Step 3: Detailed Explanation:
Given differential equation:
\[ x \frac{d^2y}{dx^2} = 1 \implies \frac{d^2y}{dx^2} = \frac{1}{x} \]
Integrating both sides with respect to $x$:
\[ \frac{dy}{dx} = \log x + c_1 \]
(We can drop the absolute value since initial conditions are at $x=1$, implying a domain of positive $x$).
Apply the first initial condition: $\frac{dy}{dx} = 0$ at $x = 1$.
\[ 0 = \log(1) + c_1 \]
Since $\log(1) = 0$, we get $c_1 = 0$.
So, the first derivative equation simplifies to:
\[ \frac{dy}{dx} = \log x \]
Now, integrate again with respect to $x$ to find $y$:
\[ y = \int \log x \,dx \]
Using integration by parts ($\int u \,dv = uv - \int v \,du$) where $u = \log x \implies du = \frac{1}{x}dx$ and $dv = dx \implies v = x$:
\[ y = x \log x - \int x \left(\frac{1}{x}\right) dx \]
\[ y = x \log x - \int 1 \,dx \]
\[ y = x \log x - x + c_2 \]
Apply the second initial condition: $y = 1$ at $x = 1$.
\[ 1 = 1 \cdot \log(1) - 1 + c_2 \]
\[ 1 = 0 - 1 + c_2 \]
\[ c_2 = 2 \]
Substitute $c_2$ back into the equation for $y$:
\[ y = x \log x - x + 2 \]
Step 4: Final Answer:
The solution is $y = x \log x - x + 2$.