Class CBSE Class 12 Mathematics Differential Equations Q #1277
KNOWLEDGE BASED
UNDERSTAND
3 Marks 2024 AISSCE(Board Exam) SA
Find the particular solution of the differential equation given by $2xy+y^{2}-2x^{2}\frac{dy}{dx}=0$ $y=2$, when $x=1.$
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Detailed Solution

Step 1: Rewrite the differential equation

The given differential equation is $2xy + y^2 - 2x^2 \frac{dy}{dx} = 0$. We can rewrite it as: $$2x^2 \frac{dy}{dx} = 2xy + y^2$$ $$\frac{dy}{dx} = \frac{2xy + y^2}{2x^2}$$

Step 2: Check for homogeneity

Let $f(x, y) = \frac{2xy + y^2}{2x^2}$. Replacing $x$ with $\lambda x$ and $y$ with $\lambda y$, we get: $$f(\lambda x, \lambda y) = \frac{2(\lambda x)(\lambda y) + (\lambda y)^2}{2(\lambda x)^2} = \frac{2\lambda^2 xy + \lambda^2 y^2}{2\lambda^2 x^2} = \frac{\lambda^2(2xy + y^2)}{\lambda^2(2x^2)} = \frac{2xy + y^2}{2x^2} = f(x, y)$$ Since $f(\lambda x, \lambda y) = f(x, y)$, the differential equation is homogeneous.

Step 3: Substitute $y = vx$

Let $y = vx$. Then, $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting these into the differential equation, we get: $$v + x\frac{dv}{dx} = \frac{2x(vx) + (vx)^2}{2x^2} = \frac{2vx^2 + v^2x^2}{2x^2} = \frac{x^2(2v + v^2)}{2x^2} = \frac{2v + v^2}{2}$$ $$x\frac{dv}{dx} = \frac{2v + v^2}{2} - v = \frac{2v + v^2 - 2v}{2} = \frac{v^2}{2}$$

Step 4: Separate variables and integrate

Separating the variables, we have: $$\frac{dv}{v^2} = \frac{dx}{2x}$$ Integrating both sides: $$\int \frac{dv}{v^2} = \int \frac{dx}{2x}$$ $$-\frac{1}{v} = \frac{1}{2} \ln|x| + C$$

Step 5: Substitute back $v = \frac{y}{x}$

Substituting $v = \frac{y}{x}$, we get: $$-\frac{1}{\frac{y}{x}} = \frac{1}{2} \ln|x| + C$$ $$-\frac{x}{y} = \frac{1}{2} \ln|x| + C$$

Step 6: Apply the initial condition $y = 2$ when $x = 1$

Substituting $x = 1$ and $y = 2$, we have: $$-\frac{1}{2} = \frac{1}{2} \ln|1| + C$$ Since $\ln(1) = 0$, we get: $$-\frac{1}{2} = 0 + C$$ $$C = -\frac{1}{2}$$

Step 7: Write the particular solution

Substituting $C = -\frac{1}{2}$ into the equation, we get: $$-\frac{x}{y} = \frac{1}{2} \ln|x| - \frac{1}{2}$$ Multiplying by -2: $$\frac{2x}{y} = 1 - \ln|x|$$ $$2x = y(1 - \ln|x|)$$ $$y = \frac{2x}{1 - \ln|x|}$$

Final Answer: $y = \frac{2x}{1 - \ln|x|}$

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the concept of homogeneous differential equations 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, and integration.
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. The question is a direct application of textbook methods.

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