Q1346: Solve the following differential equation $x^{2}dy+y(x+y)dx=...

Class CBSE Class 12 Mathematics Differential Equations Q #1346

KNOWLEDGE BASED

UNDERSTAND

3 Marks 2024 AISSCE(Board Exam) SA

Solve the following differential equation $x^{2}dy+y(x+y)dx=0$

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AI Tutor Explanation

Detailed Solution

Step 1: Rewrite the differential equation

Given the differential equation $x^{2}dy+y(x+y)dx=0$, we can rewrite it as: $$x^{2}dy = -y(x+y)dx$$ $$\frac{dy}{dx} = -\frac{y(x+y)}{x^{2}}$$ $$\frac{dy}{dx} = -\frac{xy+y^{2}}{x^{2}}$$ $$\frac{dy}{dx} = -\frac{y}{x} - \frac{y^{2}}{x^{2}}$$

Step 2: Recognize the homogeneous form

The differential equation is homogeneous because it can be written in the form $\frac{dy}{dx} = f(\frac{y}{x})$.

Step 3: Substitute $v = \frac{y}{x}$

Let $v = \frac{y}{x}$, so $y = vx$. Then, differentiate with respect to $x$: $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$

Step 4: Substitute into the differential equation

Substitute $\frac{dy}{dx}$ and $y$ into the differential equation: $$v + x\frac{dv}{dx} = -v - v^{2}$$ $$x\frac{dv}{dx} = -2v - v^{2}$$ $$x\frac{dv}{dx} = -(2v + v^{2})$$

Step 5: Separate the variables

Separate the variables: $$\frac{dv}{2v + v^{2}} = -\frac{dx}{x}$$

Step 6: Integrate both sides

Integrate both sides: $$\int \frac{dv}{2v + v^{2}} = -\int \frac{dx}{x}$$ We can rewrite the left side as: $$\int \frac{dv}{v(2 + v)} = -\int \frac{dx}{x}$$ Using partial fractions, we can write $\frac{1}{v(2+v)} = \frac{A}{v} + \frac{B}{2+v}$. $$1 = A(2+v) + Bv$$ If $v = 0$, then $1 = 2A$, so $A = \frac{1}{2}$. If $v = -2$, then $1 = -2B$, so $B = -\frac{1}{2}$. Thus, $\frac{1}{v(2+v)} = \frac{1}{2v} - \frac{1}{2(2+v)}$. Now, integrate: $$\int \left(\frac{1}{2v} - \frac{1}{2(2+v)}\right) dv = -\int \frac{dx}{x}$$ $$\frac{1}{2}\int \frac{1}{v} dv - \frac{1}{2}\int \frac{1}{2+v} dv = -\int \frac{dx}{x}$$ $$\frac{1}{2}\ln|v| - \frac{1}{2}\ln|2+v| = -\ln|x| + C_{1}$$ $$\frac{1}{2}(\ln|v| - \ln|2+v|) = -\ln|x| + C_{1}$$ $$\ln\left|\frac{v}{2+v}\right| = -2\ln|x| + 2C_{1}$$ $$\ln\left|\frac{v}{2+v}\right| = \ln|x^{-2}| + C$$ where $C = 2C_{1}$.

Step 7: Remove the logarithm

Exponentiate both sides: $$\frac{v}{2+v} = kx^{-2}$$ where $k = e^{C}$.

Step 8: Substitute back $v = \frac{y}{x}$

Substitute $v = \frac{y}{x}$: $$\frac{\frac{y}{x}}{2+\frac{y}{x}} = \frac{k}{x^{2}}$$ $$\frac{y}{2x+y} = \frac{k}{x^{2}}$$ $$yx^{2} = k(2x+y)$$ $$yx^{2} = 2kx + ky$$ $$yx^{2} - ky = 2kx$$ $$y(x^{2} - k) = 2kx$$ $$y = \frac{2kx}{x^{2} - k}$$

Step 9: Final Answer

The solution to the differential equation is: $$y = \frac{2kx}{x^{2} - k}$$

Final Answer: $y = \frac{2kx}{x^{2} - k}$

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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 substitution and integration techniques to solve it.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to apply a specific procedure (solving a homogeneous differential equation) which involves substitution, separation of variables, and integration.<\/span>

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly assesses the student's ability to solve differential equations, a standard topic in the syllabus.