Question:medium

Find the general solution of the differential equation \( \dfrac{dy}{dx} + \dfrac{y}{x} = x^2 \).

Show Hint

For first-order linear differential equations, always check the form \( \frac{dy}{dx} + P(x)y = Q(x) \). The integrating factor is \( e^{\int P(x)dx} \), which converts the left-hand side into a derivative of a product.
Updated On: Apr 15, 2026
  • \( y = \dfrac{x^3}{4} + \dfrac{C}{x} \)
  • \( y = \dfrac{x^3}{3} + \dfrac{C}{x} \)
  • \( y = \dfrac{x^3}{2} + \dfrac{C}{x} \)
  • \( y = x^3 + \dfrac{C}{x} \)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Question:
The given differential equation is a first-order linear differential equation. We need to find its general solution.
Step 2: Key Formula or Approach:
The equation is in the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \).
Here, \( P(x) = \frac{1}{x} \) and \( Q(x) = x^2 \).
The first step is to find the integrating factor (I.F.), given by \( I.F. = e^{\int P(x) dx} \).
The general solution is then given by \( y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \, dx + C \).
Step 3: Detailed Explanation:
1. Calculate the Integrating Factor (I.F.):
\[ \int P(x) dx = \int \frac{1}{x} dx = \ln|x| \] \[ I.F. = e^{\ln|x|} = |x| \] Since we are looking for a general solution, we can take \( I.F. = x \).
2. Apply the solution formula:
\[ y \cdot x = \int x^2 \cdot x \, dx + C \] \[ yx = \int x^3 \, dx + C \] 3. Integrate the right side:
\[ yx = \frac{x^4}{4} + C \] 4. Solve for y:
Divide the entire equation by \( x \):
\[ y = \frac{x^3}{4} + \frac{C}{x} \] Step 4: Final Answer:
The general solution of the differential equation is \( y = \dfrac{x^3}{4} + \dfrac{C}{x} \).
Was this answer helpful?
0