CBSE Class 12 Mathematics Differential Equations Q #2093
COMPETENCY BASED
APPLY
3 Marks 2026-27 CBSE Sample QP SA
Find the particular solution of the differential equation $x dy + (y - x^2 e^x) dx = 0$, given that $y(1)=0$
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Detailed Solution

Step 1: Standardize the Differential Equation

The given equation is $x dy + (y - x^2 e^x) dx = 0$. Rearranging it to the form $\frac{dy}{dx} + P(x)y = Q(x)$, we get: $$x \frac{dy}{dx} + y = x^2 e^x$$ Dividing by $x$ (assuming $x \neq 0$): $$\frac{dy}{dx} + \frac{1}{x}y = x e^x$$

Step 2: Find the Integrating Factor (IF)

Here, $P(x) = \frac{1}{x}$ and $Q(x) = x e^x$. The integrating factor is given by: $$IF = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$$

Step 3: Solve the Equation

The general solution is $y \cdot IF = \int (Q(x) \cdot IF) dx + C$: $$y \cdot x = \int (x e^x \cdot x) dx = \int x^2 e^x dx$$ Using integration by parts ($\int u dv = uv - \int v du$): $$\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx = x^2 e^x - 2(x e^x - e^x) + C$$ $$xy = x^2 e^x - 2x e^x + 2e^x + C$$

Step 4: Apply Initial Conditions

Given $y(1) = 0$, substitute $x=1$ and $y=0$: $$0 \cdot 1 = (1)^2 e^1 - 2(1) e^1 + 2e^1 + C$$ $$0 = e - 2e + 2e + C \implies C = -e$$ Thus, $xy = e^x(x^2 - 2x + 2) - e$.

Final Answer: y = \frac{e^x(x^2 - 2x + 2) - e}{x}

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to identify the linear differential equation form and execute a multi-step algorithmic procedure to find the particular solution.
Knowledge Dimension: PROCEDURAL
Justification: The student must follow a specific sequence of mathematical operations: standardizing the equation, calculating the IF, performing integration by parts, and solving for the constant.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the student's ability to handle linear differential equations, a core topic in the Calculus unit, requiring both algebraic manipulation and integration techniques.

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