Variable Separable
Solve the differential equation \frac{dy}{dx} = e^{x-y} and express the general solution relating the exponentials of x and y explicitly with one arbitrary constant.
Select the correct option:
Solution
ey=ex+C
The exponential e^{x-y} factors as e^x \cdot e^{-y}, which immediately reveals a separable structure because the x-dependence and y-dependence split as a product. The strategic step is to recognize that any exponential of a sum or difference of variables decomposes into separable factors. Rewriting the equation, \frac{dy}{dx} = e^x e^{-y}, and separating gives e^{y},dy = e^{x},dx. Integrating both sides yields e^{y} = e^{x} + C, where C is the single arbitrary constant. Option e^{-y} = e^x + C arises from forgetting to move the negative exponential to the left correctly, which would flip the sign of the y exponent. Option y = e^x + C omits the exponential on the left entirely, treating dy as though it integrated to y rather than e^y,dy integrating to e^y. Option e^y = e^{-x} + C reverses the sign of the x exponent and contradicts the separation. This is the textbook JEE Advanced separable-exponential pattern. As a final plausibility check, differentiating e^y = e^x + C implicitly gives e^y y' = e^x, so y' = e^x e^{-y} = e^{x-y}, exactly the original equation, confirming the solution is correct.
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About This Question
- Subject
- mathematics
- Chapter
- differential equations
- Topic
- variable separable
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
ey=ex+C
The exponential e^{x-y} factors as e^x \cdot e^{-y}, which immediately reveals a separable structure because the x-dependence and y-dependence split as a product. The strategic step is to recognize that any exponential of a sum or difference of variables decomposes into separable factors. Rewriting the equation, \frac{dy}{dx} = e^x e^{-y}, and separating gives e^{y},dy = e^{x},dx. Integrating both sides yields e^{y} = e^{x} + C, where C is the single arbitrary constant. Option e^{-y} = e^x + C arises from forgetting to move the negative exponential to the left correctly, which would flip the sign of the y exponent. Option y = e^x + C omits the exponential on the left entirely, treating dy as though it integrated to y rather than e^y,dy integrating to e^y. Option e^y = e^{-x} + C reverses the sign of the x exponent and contradicts the separation. This is the textbook JEE Advanced separable-exponential pattern. As a final plausibility check, differentiating e^y = e^x + C implicitly gives e^y y' = e^x, so y' = e^x e^{-y} = e^{x-y}, exactly the original equation, confirming the solution is correct.
This easy difficulty mathematics question is from the chapter differential equations, covering the topic of variable separable. It appeared in the 2025 exam.
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