Question

The solution of the differential equation \(x \frac{d^{2} y}{dx^{2}} + \frac{dy}{dx} = 0\) is

• A. \(y = c_{1} \sin x + c_{2}\)
• B. \(y = c_{1} \cos x + c_{2}\)
• C. \(y = c_{1} \log x + c_{2}\)
• D. \(y = c_{1} e^{x} + c_{2} e^{-x}\)

Hint:

Notice that the equation can be written using the product rule:
$x \frac{d^2 y}{dx^2} + \frac{dy}{dx} = \frac{d}{dx}\left( x \frac{dy}{dx} \right) = 0$.
Integrating once gives $x \frac{dy}{dx} = c_1$, and integrating a second time leads directly to $y = c_1 \ln x + c_2$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the general solution to a second-order, linear, homogeneous ordinary differential equation.

Step 2: Key Formula or Approach:
This equation can be solved by reducing its order. Let \(p = \frac{dy}{dx}\). Then the second derivative becomes \(\frac{d^{2} y}{dx^{2}} = \frac{dp}{dx}\).
Substitute these into the equation to obtain a first-order separable differential equation in terms of \(p\) and \(x\).

Step 3: Detailed Explanation:

• Substitute \(p\) and \(\frac{dp}{dx}\) into the original differential equation:
\[ x \frac{dp}{dx} + p = 0 \]
• Separate the variables:
\[ x \frac{dp}{dx} = -p \implies \frac{dp}{p} = -\frac{dx}{x} \]
• Integrate both sides of the equation:
\[ \int \frac{dp}{p} = -\int \frac{dx}{x} \] \[ \ln p = -\ln x + \ln c_{1} \] \[ \ln p = \ln\left(\frac{c_{1}}{x}\right) \implies p = \frac{c_{1}}{x} \]
• Replace \(p\) with \(\frac{dy}{dx}\) to solve for \(y\):
\[ \frac{dy}{dx} = \frac{c_{1}}{x} \]
• Integrate with respect to \(x\):
\[ y = \int \frac{c_{1}}{x} dx \] \[ y = c_{1} \ln x + c_{2} \] Since \(\ln x\) and \(\log x\) are equivalent in standard calculus notation:
\[ y = c_{1} \log x + c_{2} \]

Step 4: Final Answer:
The general solution to the differential equation is \(y = c_{1} \log x + c_{2}\).