Question

Particular solution of the differential equation \[ \frac{dy}{dx}+2y^{2}=0,\quad y(1)=1 \] is:

• A. \(y=2x-1\)
• B. \(y=1-2x\)
• C. \(y=\frac{1}{2x-1} \)
• D. \(y=\frac{1}{1-2x} \)

Hint:

For equations of type $y' = ky^2$, the solution is always a reciprocal linear function.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a separable differential equation: \( \frac{dy}{dx} = -2y^2 \).

Step 2: Detailed Explanation:
Separate variables: \( \frac{dy}{y^2} = -2 dx \).
Integrate: \( \int y^{-2} dy = \int -2 dx \implies -y^{-1} = -2x + c \implies \frac{1}{y} = 2x - c \).
Use initial condition \( y=1 \) when \( x=1 \):
\( \frac{1}{1} = 2(1) - c \implies 1 = 2 - c \implies c = 1 \).
The equation becomes \( \frac{1}{y} = 2x - 1 \implies y = \frac{1}{2x - 1} \).

Step 3: Final Answer:
The particular solution is \( y = \frac{1}{2x - 1} \), which is option (C).