Class CBSE Class 12 Mathematics Differential Equations Q #1715
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
The general solution of the differential equation $\frac{dy}{dx}=\frac{\sqrt{y}}{\sqrt{x}}$ is
(A) $\log\sqrt{y}=\log\sqrt{x}+C$
(B) $\sqrt{y}+\sqrt{x}=C$
(C) $\sqrt{y}-\sqrt{x}=C$
(D) $\log\sqrt{y}+\log\sqrt{x}=C$
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Correct Answer: C

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Detailed Solution

Step 1: Separate the variables

The given differential equation is $$ \frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}} $$ Separating the variables $x$ and $y$, we get: $$ \frac{1}{\sqrt{y}} dy = \frac{1}{\sqrt{x}} dx $$ Which can be written as: $$ y^{-1/2} dy = x^{-1/2} dx $$

Step 2: Integrate both sides

Integrating both sides: $$ \int y^{-1/2} dy = \int x^{-1/2} dx $$ Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$: $$ \frac{y^{1/2}}{1/2} = \frac{x^{1/2}}{1/2} + C $$ $$ 2\sqrt{y} = 2\sqrt{x} + C $$

Step 3: Simplify the expression

Dividing the entire equation by 2: $$ \sqrt{y} = \sqrt{x} + \frac{C}{2} $$ Let $C_1 = \frac{C}{2}$, then: $$ \sqrt{y} - \sqrt{x} = C_1 $$ This matches option (C).

Final Answer: $\sqrt{y}-\sqrt{x}=C$

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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 standard integration techniques to solve a differential equation.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the execution of a specific algorithmic process (separation of variables followed by integration) taught in the CBSE Class 12 curriculum.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question tests the student's ability to manipulate algebraic expressions and apply calculus concepts to solve differential equations, which is a core competency in the 'Calculus' unit.

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