Class CBSE Class 12 Mathematics Differential Equations Q #1322
COMPETENCY BASED
UNDERSTAND
3 Marks 2024 AISSCE(Board Exam) SA
Find the particular solution of the differential equation $(xe^{\frac{y}{x}}+y)dx=x~dy$, given that $y=1$ when $x=1$
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Detailed Solution

Step 1: Rewrite the differential equation

Given the differential equation $(xe^{\frac{y}{x}}+y)dx=x~dy$, we can rewrite it as: $$x\frac{dy}{dx} = xe^{\frac{y}{x}} + y$$ $$\frac{dy}{dx} = e^{\frac{y}{x}} + \frac{y}{x}$$

Step 2: Recognize the homogeneous form

The equation is homogeneous. Let $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting these into the equation, we get: $$v + x\frac{dv}{dx} = e^v + v$$ $$x\frac{dv}{dx} = e^v$$

Step 3: Separate variables and integrate

Separate the variables: $$\frac{dv}{e^v} = \frac{dx}{x}$$ $$e^{-v} dv = \frac{dx}{x}$$ Integrate both sides: $$\int e^{-v} dv = \int \frac{dx}{x}$$ $$-e^{-v} = \ln|x| + C$$

Step 4: Substitute back for v

Substitute $v = \frac{y}{x}$ back into the equation: $$-e^{-\frac{y}{x}} = \ln|x| + C$$

Step 5: Apply the initial condition

Given $y = 1$ when $x = 1$, substitute these values into the equation: $$-e^{-\frac{1}{1}} = \ln|1| + C$$ $$-e^{-1} = 0 + C$$ $$C = -e^{-1} = -\frac{1}{e}$$

Step 6: Write the particular solution

Substitute the value of $C$ back into the equation: $$-e^{-\frac{y}{x}} = \ln|x| - \frac{1}{e}$$ $$e^{-\frac{y}{x}} = \frac{1}{e} - \ln|x|$$ $$-\frac{y}{x} = \ln\left(\frac{1}{e} - \ln|x|\right)$$ $$y = -x \ln\left(\frac{1}{e} - \ln|x|\right)$$ $$y = -x \ln\left(\frac{1 - e\ln|x|}{e}\right)$$ $$y = -x \left[ \ln(1 - e\ln|x|) - \ln(e) \right]$$ $$y = -x \left[ \ln(1 - e\ln|x|) - 1 \right]$$ $$y = x - x\ln(1 - e\ln|x|)$$

Final Answer: $y = x - x\ln(1 - e\ln|x|)$

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because it requires the student to recognize the type of differential equation and apply the appropriate method to solve it.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific procedure (solving a homogeneous differential equation) to arrive at the solution. This involves steps like substitution, separation of variables, integration, and applying initial conditions.
Syllabus Audit: In the context of CBSE Class 12, this is classified as APPLICATION. The question requires the student to apply the concepts of differential equations and their solutions to find a particular solution given an initial condition.

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