Question

Solution of \( (2y - x) \frac{dy}{dx} = 1 \) is

• A. \( x = 2(y - 1) + ce^{-y} \), where c is the constant of integration
• B. \( x = 2(y - 1) + ce^{-x} \), where c is the constant of integration
• C. \( y = 2(x - 1) + ce^{-x} \), where c is the constant of integration
• D. \( y = 2(x - 1) + ce^{-y} \), where c is the constant of integration

Hint:

If $dy/dx$ looks complicated, try inverted form $dx/dy$; it often turns into a standard linear DE.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This equation is not directly integrable as a linear differential equation in \( y \). Let's check if it's linear in \( x \).
Step 2: Key Formula or Approach:
1. Rewrite as \( \frac{dx}{dy} = 2y - x \implies \frac{dx}{dy} + x = 2y \).
2. This is a linear differential equation of form \( \frac{dx}{dy} + P(y)x = Q(y) \).
3. Integrating factor \( IF = e^{\int P(y) dy} \).
4. Solution: \( x \cdot IF = \int Q(y) \cdot IF dy + C \).
Step 3: Detailed Explanation:
Here \( P(y) = 1 \) and \( Q(y) = 2y \).
\( IF = e^{\int 1 dy} = e^y \).
The solution is:
\[ x e^y = \int 2y e^y dy + c \] Using integration by parts \( \int u v dy = u \int v dy - \int (u' \int v dy) dy \):
\[ x e^y = 2 [y e^y - \int e^y dy] + c = 2 [y e^y - e^y] + c = 2e^y(y - 1) + c \] Divide by \( e^y \):
\[ x = 2(y - 1) + c e^{-y} \] Step 4: Final Answer:
The solution is \( x = 2(y - 1) + ce^{-y} \).