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: Verify each option by differentiation:
Instead of solving the equation from scratch, differentiate each given relation implicitly and check which one gives back $\dfrac{dy}{dx}=e^{x+y}$.

Step 2: Test option B, $e^{x}+e^{-y}=c$:
Differentiate both sides with respect to x.
\[ e^{x} + e^{-y}\left(-\dfrac{dy}{dx}\right) = 0 \]
\[ \dfrac{dy}{dx} = \dfrac{e^{x}}{e^{-y}} = e^{x}\cdot e^{y} = e^{x+y} \]
This exactly matches the given differential equation.

Step 3: Confirm no other option works:
Repeating the same differentiation test on options A, C and D produces expressions like $e^{x+y}$ with a wrong sign, or $e^{y-x}$, none of which equal $e^{x+y}$ exactly.
So only the relation in option B survives the check.

Step 4: Cross check by the constant of integration:
Since the original separable form gives $\int e^{-y}dy=\int e^{x}dx$, the natural family of solutions is $-e^{-y}=e^{x}+c_1$, which rearranges to the same relation $e^{x}+e^{-y}=c$.

Final Answer:
Differentiating option B reproduces the given equation exactly, confirming it is correct. \[ \boxed{e^{x}+e^{-y}=c} \]