Particular solution of the differential equation \( dy=e^{2x+y}dx \), \( y(0)=0 \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( y=\log\left|\displaystyle\frac{2}{3-e^{2x}}\right| \)
Since \( e^{2x+y}=e^{2x}e^{y} \), the variables separate:
\( \displaystyle e^{-y}\,dy=e^{2x}\,dx \)
\( \displaystyle \int e^{-y}\,dy=\int e^{2x}\,dx \)
\( \displaystyle -e^{-y}=\frac{e^{2x}}{2}+\mathrm{C} \)
Putting x = 0, y = 0: \( -1=\frac{1}{2}+\mathrm{C} \), so \( \mathrm{C}=-\frac{3}{2} \).
\( \displaystyle -e^{-y}=\frac{e^{2x}}{2}-\frac{3}{2} \)
\( \displaystyle e^{-y}=\frac{3-e^{2x}}{2} \)
\( \displaystyle e^{y}=\frac{2}{3-e^{2x}} \)
Taking logarithms,
\( \displaystyle y=\log\left|\frac{2}{3-e^{2x}}\right| \)
Shortcut: Check y(0) = 0 in the options: Option (1) gives log 1 = 0; Option (2) gives log 2; Option (3) gives 1; Option (4) gives 2. Only Option (1) satisfies the initial condition.