Class CBSE Class 12 Mathematics Differential Equations Q #1384
KNOWLEDGE BASED
APPLY
3 Marks 2025 AISSCE(Board Exam) SA
Solve the differential equation $2(y+3)-xy\frac{dy}{dx}=0;$ given $y(1)=-2$.
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Detailed Solution

Step 1: Rearrange the differential equation

We are given the differential equation $2(y+3)-xy\frac{dy}{dx}=0$. We can rearrange this equation to separate the variables $x$ and $y$. $2(y+3) = xy\frac{dy}{dx}$ $\frac{2(y+3)}{y} dy = x dx$ $\frac{2y+6}{y} dy = x dx$ $(2 + \frac{6}{y}) dy = x dx$

Step 2: Integrate both sides

Now, we integrate both sides of the equation with respect to their respective variables: $\int (2 + \frac{6}{y}) dy = \int x dx$ $2\int dy + 6\int \frac{1}{y} dy = \int x dx$ $2y + 6\ln|y| = \frac{x^2}{2} + C$, where $C$ is the constant of integration.

Step 3: Apply the initial condition

We are given the initial condition $y(1) = -2$. We substitute $x=1$ and $y=-2$ into the equation to find the value of $C$: $2(-2) + 6\ln|-2| = \frac{(1)^2}{2} + C$ $-4 + 6\ln(2) = \frac{1}{2} + C$ $C = -4 + 6\ln(2) - \frac{1}{2}$ $C = -\frac{9}{2} + 6\ln(2)$

Step 4: Write the final solution

Substitute the value of $C$ back into the equation: $2y + 6\ln|y| = \frac{x^2}{2} - \frac{9}{2} + 6\ln(2)$ Multiply by 2: $4y + 12\ln|y| = x^2 - 9 + 12\ln(2)$ $4y - x^2 + 12\ln|y| - 12\ln(2) + 9 = 0$ $4y - x^2 + 12\ln|\frac{y}{2}| + 9 = 0$

Final Answer: $4y - x^2 + 12\ln|\frac{y}{2}| + 9 = 0$

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the concepts of differential equations, separation of variables, integration, and initial value problems to find the particular solution.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure to solve the differential equation, including separating variables, integrating, and applying the initial condition.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of solving differential equations, a standard topic in the syllabus.

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