Let \( y(x) \) be the solution curve of the differential equation \( \displaystyle\frac{dy}{dx}+2y^2=0,y(1)=1 \) , then \( y\left(\displaystyle\frac{3}{4}\right) \) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
The variables separate:
\( \displaystyle \frac{dy}{dx}=-2y^2 \)
\( \displaystyle \frac{dy}{y^2}=-2\,dx \)
\( \displaystyle -\frac{1}{y}=-2x+c \)
Putting x = 1, y = 1:
\( \displaystyle -1=-2+c\ \Rightarrow\ c=1 \)
\( \displaystyle -\frac{1}{y}=-2x+1 \)
\( \displaystyle y=\frac{1}{2x-1} \)
Putting \( x=\frac{3}{4} \):
\( \displaystyle y\left(\frac{3}{4}\right)=\frac{1}{\frac{3}{2}-1} \)
\( \displaystyle =\frac{1}{\frac{1}{2}} \)
\( \displaystyle =2 \)