Q648: The differential equation $\frac{dy}{dx}=F(x,y)$ will not ...

Class CBSE Class 12 Mathematics Differential Equations Q #648

KNOWLEDGE BASED

APPLY

1 Marks 2024 AISSCE(Board Exam) MCQ SINGLE

The differential equation $\frac{dy}{dx}=F(x,y)$ will not be a homogeneous differential equation, if $F(x,y)$ is :

(A) $\cos~x-\sin(\frac{y}{x})$

(B) $\frac{y}{x}$

(D) $\cos^{2}(\frac{x}{y})$

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Correct Answer: A

AI Tutor Explanation

Step-by-Step Solution

A function $F(x, y)$ is homogeneous of degree $n$ if $F(tx, ty) = t^n F(x, y)$ for some constant $n$.

Let's check each option:

(A) $F(x, y) = \cos x - \sin(\frac{y}{x})$

$F(tx, ty) = \cos(tx) - \sin(\frac{ty}{tx}) = \cos(tx) - \sin(\frac{y}{x})$. This is not of the form $t^n F(x, y)$ because of the $\cos(tx)$ term. Therefore, this is not a homogeneous function.

(B) $F(x, y) = \frac{y}{x}$

$F(tx, ty) = \frac{ty}{tx} = \frac{y}{x} = t^0 \frac{y}{x}$. This is homogeneous of degree 0.

$F(tx, ty) = \frac{(tx)^2 + (ty)^2}{(tx)(ty)} = \frac{t^2x^2 + t^2y^2}{t^2xy} = \frac{t^2(x^2 + y^2)}{t^2xy} = \frac{x^2 + y^2}{xy} = t^0 \frac{x^2 + y^2}{xy}$. This is homogeneous of degree 0.

(D) $F(x, y) = \cos^2(\frac{x}{y})$

$F(tx, ty) = \cos^2(\frac{tx}{ty}) = \cos^2(\frac{x}{y}) = t^0 \cos^2(\frac{x}{y})$. This is homogeneous of degree 0.

Therefore, the function that is not homogeneous is option (A).

Correct Answer: cos~x-sin(y/x)

APPLY|||QUESTION|||CONCEPTUAL|||MEDIUM|||

Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student needs to apply the definition of a homogeneous differential equation to determine which of the given functions does not satisfy the homogeneity condition.

Knowledge Dimension: CONCEPTUAL

Justification: The question requires understanding the concept of homogeneous differential equations and applying the definition to identify non-homogeneous functions.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the understanding and application of the definition of homogeneous differential equations, a standard topic in the syllabus.

Step-by-Step Solution

A function $F(x, y)$ is homogeneous of degree $n$ if $F(tx, ty) = t^n F(x, y)$ for some constant $n$.

Let's check each option:

(A) $F(x, y) = \cos x - \sin(\frac{y}{x})$

$F(tx, ty) = \cos(tx) - \sin(\frac{ty}{tx}) = \cos(tx) - \sin(\frac{y}{x})$. This is not of the form $t^n F(x, y)$ because of the $\cos(tx)$ term. Therefore, this is not a homogeneous function.

(B) $F(x, y) = \frac{y}{x}$

$F(tx, ty) = \frac{ty}{tx} = \frac{y}{x} = t^0 \frac{y}{x}$. This is homogeneous of degree 0.

$F(tx, ty) = \frac{(tx)^2 + (ty)^2}{(tx)(ty)} = \frac{t^2x^2 + t^2y^2}{t^2xy} = \frac{t^2(x^2 + y^2)}{t^2xy} = \frac{x^2 + y^2}{xy} = t^0 \frac{x^2 + y^2}{xy}$. This is homogeneous of degree 0.

(D) $F(x, y) = \cos^2(\frac{x}{y})$

$F(tx, ty) = \cos^2(\frac{tx}{ty}) = \cos^2(\frac{x}{y}) = t^0 \cos^2(\frac{x}{y})$. This is homogeneous of degree 0.

Therefore, the function that is not homogeneous is option (A).

Correct Answer: cos~x-sin(y/x)

AI Suggestion: Option A

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Pedagogical Audit

Knowledge Dimension: CONCEPTUAL

Justification: The question requires understanding the concept of homogeneous differential equations and applying the definition to identify non-homogeneous functions.

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