Class CBSE Class 12 Mathematics Differential Equations Q #1506
COMPETENCY BASED
APPLY
3 Marks 2026 AISSCE(Board Exam) SA
Find the particular solution of the differential equation $x\frac{dy}{dx}=(x+2)(y+2)$, given that $y(1)=-1$.
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Detailed Solution

Step 1: Separate the variables

The given differential equation is $x\frac{dy}{dx} = (x+2)(y+2)$. We rearrange the terms to separate $x$ and $y$ variables:

$$\frac{dy}{y+2} = \frac{x+2}{x} dx$$ $$\frac{dy}{y+2} = (1 + \frac{2}{x}) dx$$

Step 2: Integrate both sides

Integrating both sides with respect to their variables:

$$\int \frac{1}{y+2} dy = \int (1 + \frac{2}{x}) dx$$ $$\ln|y+2| = x + 2\ln|x| + C$$

Step 3: Apply the initial condition

Given $y(1) = -1$, substitute $x=1$ and $y=-1$ into the equation to find $C$:

$$\ln|-1+2| = 1 + 2\ln|1| + C$$ $$\ln(1) = 1 + 0 + C$$ $$0 = 1 + C \implies C = -1$$

Step 4: Write the final solution

Substitute $C = -1$ back into the general solution:

$$\ln|y+2| = x + 2\ln|x| - 1$$ $$\ln|y+2| - \ln(x^2) = x - 1$$ $$\ln|\frac{y+2}{x^2}| = x - 1$$ $$\frac{y+2}{x^2} = e^{x-1} \implies y = x^2 e^{x-1} - 2$$

Final Answer: y = x^2 e^{x-1} - 2

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must identify the method of separation of variables and apply integration techniques to solve a specific initial value problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a sequential execution of mathematical steps (separation, integration, constant evaluation) to reach the solution.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. This aligns with the NCERT curriculum for Differential Equations, testing the student's ability to handle non-homogeneous separable equations.

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