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

Concept: This is a separable differential equation of the form $\frac{dy}{dx}=f(y)$ which can be solved by separating variables and integrating.

Step 1: {Rewrite equation.}
\[ \frac{dy}{dx}=-2y^{2} \]

Step 2: {Separate variables.}
\[ \frac{dy}{y^{2}}=-2dx \]

Step 3: {Integrate both sides.}
\[ \int y^{-2}dy=\int -2dx \]

Step 4: {Solve integrals.}
\[ -\frac{1}{y}=-2x+c \]

Step 5: {Rearrange.}
\[ \frac{1}{y}=2x+c \]

Step 6: {Apply initial condition $y(1)=1$.}
\[ 1=2+c \Rightarrow c=-1 \]

Step 7: {Final solution.}
\[ y=\frac{1}{1-2x} \]