The particular solution of the differential equation \( dy=e^{2x+y}dx \), \( y(0)=0 \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( y=\log_e\left|\displaystyle\frac{2}{3-e^{2x}}\right| \)
The equation is in variables separable form, since \( e^{2x+y}=e^{2x}\cdot e^{y} \):
\( \displaystyle dy=e^{2x}e^{y}\,dx \)
\( \displaystyle e^{-y}\,dy=e^{2x}\,dx \)
Integrating both sides:
\( \displaystyle \int e^{-y}\,dy=\int e^{2x}\,dx \)
\( \displaystyle -e^{-y}=\frac{e^{2x}}{2}+C \)
Putting x = 0, y = 0:
\( \displaystyle -1=\frac{1}{2}+C\quad\Rightarrow\quad C=-\frac{3}{2} \)
So
\( \displaystyle -e^{-y}=\frac{e^{2x}}{2}-\frac{3}{2} \)
\( \displaystyle e^{-y}=\frac{3-e^{2x}}{2} \)
\( \displaystyle e^{y}=\frac{2}{3-e^{2x}} \)
\( \displaystyle y=\log_e\left|\frac{2}{3-e^{2x}}\right| \)
Why the other options are incorrect
Option (1) \( x=\log_e\left|\displaystyle\frac{3+e^{2y}}{2}\right| \) — Incorrect. At x = 0, y = 0 this gives \( 0=\log_e 2 \), so it does not satisfy the initial condition.
Option (2) \( x=\log_e\left|\displaystyle\frac{3-e^{2y}}{2}\right| \) — Incorrect. It passes through (0, 0), but it is not a solution: differentiating \( e^{x}=\frac{3-e^{2y}}{2} \) gives \( e^{x}dx=-e^{2y}dy \), which is not the given equation.
Option (4) \( y=\log_e\left|\displaystyle\frac{2}{3+e^{2x}}\right| \) — Incorrect. At x = 0 it gives \( y=\log_e\frac{1}{2}\ne 0 \), so the initial condition fails.