Class CBSE Class 12 Mathematics Differential Equations Q #1717
KNOWLEDGE BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
The general solution for the differential equation $\frac{dy}{dx}=e^{3x-y}$ is:
(A) $3e^{y}=e^{3x}+C$
(B) $\log(3x-y)=C$
(C) $e^{3x-y}=C$
(D) $-e^{y}+3e^{3x}=C$
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Correct Answer: A

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Detailed Solution

Step 1: Separate the variables

The given differential equation is $$ \frac{dy}{dx} = e^{3x} \cdot e^{-y} $$ By separating the variables $x$ and $y$, we get: $$ e^{y} \, dy = e^{3x} \, dx $$

Step 2: Integrate both sides

Integrate both sides of the equation: $$ \int e^{y} \, dy = \int e^{3x} \, dx $$ The integral of $e^{y}$ is $e^{y}$ and the integral of $e^{3x}$ is $\frac{e^{3x}}{3}$. Adding the constant of integration $C$: $$ e^{y} = \frac{e^{3x}}{3} + C $$

Step 3: Simplify the expression

To match the given options, multiply the entire equation by 3: $$ 3e^{y} = e^{3x} + 3C $$ Since $3C$ is still an arbitrary constant, we can replace it with $C$: $$ 3e^{y} = e^{3x} + C $$

Final Answer: $3e^{y}=e^{3x}+C$

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must apply the method of separation of variables and integration techniques to solve a standard differential equation.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the execution of a specific sequence of mathematical steps (separation, integration, and algebraic manipulation) to reach the solution.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. This is a direct application of the "Differential Equations" chapter, specifically the "Variable Separable" method, which is a core component of the NCERT curriculum.

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