Question

If \(\dfrac{dy}{dx}=e^{x+y}\), then its solution is:

• A. \(e^{-y}=e^{x}+c\)
• B. \(e^{x}+e^{-y}=c\)
• C. \(e^{-x}-e^{-y}=c\)
• D. \(e^{-x}+e^{-y}=c\)

Hint:

Write \(e^{x+y}=e^x e^y\), separate variables as \(e^{-y}dy=e^x dx\), then integrate.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The equation \( \dfrac{dy}{dx}=e^{x+y} \) can be split using the law of exponents \( e^{x+y}=e^{x}\cdot e^{y} \).
Once split this way, all the x terms can be moved to one side and all the y terms to the other, so the equation is variable separable.

Step 2: Separating the variables:
Write \( \dfrac{dy}{dx}=e^{x}\cdot e^{y} \).
Divide both sides by \( e^{y} \) so that y terms stay on the left and x terms move to the right.
\[ e^{-y}\,dy = e^{x}\,dx \]

Step 3: Integrating both sides:
Integrate the left side with respect to y and the right side with respect to x.
\[ \int e^{-y}\,dy = \int e^{x}\,dx \]
\[ -e^{-y} = e^{x} + c_1 \]
Multiply through by \(-1\) and rename the constant to get the final relation.
\[ e^{-y} = -e^{x} - c_1 \quad\Rightarrow\quad e^{x}+e^{-y}=c \]

Step 4: Why option A is wrong:
Option A, \( e^{-y}=e^{x}+c \), drops the sign that comes from integrating \( e^{-y} \); it does not satisfy the original equation on differentiation.

Step 5: Why option C is wrong:
Option C, \( e^{-x}-e^{-y}=c \), differentiates to \( -e^{-x}+e^{-y}\dfrac{dy}{dx}=0 \), giving \( \dfrac{dy}{dx}=e^{y-x} \), not \( e^{x+y} \).

Step 6: Why option D is wrong:
Option D, \( e^{-x}+e^{-y}=c \), differentiates to \( -e^{-x}-e^{-y}\dfrac{dy}{dx}=0 \), giving \( \dfrac{dy}{dx}=-e^{y+x} \), which has the wrong sign.

Final Answer:
Only \( e^{x}+e^{-y}=c \) reproduces the given equation on differentiation. \[ \boxed{e^{x}+e^{-y}=c} \]