Class CBSE Class 12 Mathematics Differential Equations Q #1493
KNOWLEDGE BASED
REMEMBER
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) $$-e^y + 3e^{3x} = C$$
Prev Finish
Correct Answer: A

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Detailed Solution

Step 1: Separate the variables

Given the differential equation $\frac{dy}{dx} = e^{3x-y}$, we can rewrite it as $\frac{dy}{dx} = e^{3x} \cdot e^{-y}$. Separating the variables, we get $e^y dy = e^{3x} dx$.

Step 2: Integrate both sides

Integrating both sides of the equation $e^y dy = e^{3x} dx$, we have $\int e^y dy = \int e^{3x} dx$.

Step 3: Evaluate the integrals

The integral of $e^y$ with respect to $y$ is $e^y$. The integral of $e^{3x}$ with respect to $x$ is $\frac{1}{3}e^{3x}$. Therefore, we have $e^y = \frac{1}{3}e^{3x} + C_1$, where $C_1$ is the constant of integration.

Step 4: Rearrange the equation

Multiplying both sides by 3, we get $3e^y = e^{3x} + 3C_1$. Let $C = 3C_1$, then $3e^y = e^{3x} + C$.

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

AI Suggestion: Option A

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because the student needs to recall the method of solving differential equations by separating variables and integrating.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific procedure (separation of variables and integration) to solve the differential equation.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding of differential equations and their solutions, a standard topic in the syllabus.
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