Question

The solution of the differential equation \(xy^{\prime\prime}+y^{\prime}=4x\) is given by:

• A. $\text{y} = \text{x}^3 + \text{C}_1 \ln\text{x} + \text{C}_2$
• B. $\text{y} = \text{x}^2 + \text{C}_1 \ln\text{x} + \text{C}_2$
• C. $\text{y} = \text{e}^{\text{x}} + \text{C}_1 \ln\text{x} + \text{C}_2$
• D. $\text{y} = \text{e}^{-\text{x}} + \text{C}_1 \ln\text{x} + \text{C}_2$

Hint:

Recognizing the product rule on the LHS, $x y'' + y' = \frac{d}{dx}(x y')$, is a standard shortcut for solving equations of this specific form.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a second-order ordinary differential equation.
Since the dependent variable $y$ is missing from the equation, we can reduce its order by introducing a substitution for the first derivative.
Key Formula or Approach:
Let $p = y' = \frac{dy}{dx}$.
Then $y'' = \frac{dp}{dx}$.
The differential equation reduces to a first-order linear differential equation in terms of $p$.

Step 2: Detailed Explanation:
Substitute $y'$ and $y''$ into the differential equation: \[ x \frac{dp}{dx} + p = 4x \] Notice that the left-hand side is the derivative of the product $xp$: \[ \frac{d}{dx}(xp) = 4x \] Integrate both sides with respect to $x$: \[ xp = \int 4x \, dx \] \[ xp = 2x^2 + C_1 \] Divide by $x$: \[ p = 2x + \frac{C_1}{x} \] Since $p = \frac{dy}{dx}$: \[ \frac{dy}{dx} = 2x + \frac{C_1}{x} \] Integrate with respect to $x$ to find $y$: \[ y = \int \left( 2x + \frac{C_1}{x} \right) dx \] \[ y = x^2 + C_1 \ln x + C_2 \] This is the general solution of the differential equation.

Step 3: Final Answer:
The solution matches Option (B).