Class CBSE Class 12 Mathematics Differential Equations Q #1321
KNOWLEDGE BASED
REMEMBER
3 Marks 2024 AISSCE(Board Exam) SA
Find the particular solution of the differential equation $\frac{dy}{dx}=y~cot~2x,$ given that $y(\frac{\pi}{4})=2.$

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Detailed Solution

Step 1: Separate the variables

We are given the differential equation $\frac{dy}{dx} = y \cot 2x$. To solve this, we first separate the variables $y$ and $x$ to get: $$\frac{dy}{y} = \cot 2x ~dx$$

Step 2: Integrate both sides

Now, we integrate both sides of the equation: $$\int \frac{dy}{y} = \int \cot 2x ~dx$$ The integral of $\frac{1}{y}$ with respect to $y$ is $\ln |y|$. The integral of $\cot 2x$ with respect to $x$ is $\frac{1}{2} \ln |\sin 2x|$. Thus, we have: $$\ln |y| = \frac{1}{2} \ln |\sin 2x| + C$$ where $C$ is the constant of integration.

Step 3: Simplify the equation

We can rewrite the equation as: $$\ln |y| = \ln |\sqrt{\sin 2x}| + C$$ Taking the exponential of both sides, we get: $$|y| = e^{\ln |\sqrt{\sin 2x}| + C} = e^{\ln |\sqrt{\sin 2x}|} \cdot e^C$$ $$y = A \sqrt{\sin 2x}$$ where $A = \pm e^C$ is another constant.

Step 4: Apply the initial condition

We are given that $y(\frac{\pi}{4}) = 2$. Substituting $x = \frac{\pi}{4}$ and $y = 2$ into the general solution, we get: $$2 = A \sqrt{\sin (2 \cdot \frac{\pi}{4})} = A \sqrt{\sin (\frac{\pi}{2})} = A \sqrt{1} = A$$ Thus, $A = 2$.

Step 5: Write the particular solution

Substituting $A = 2$ into the general solution, we obtain the particular solution: $$y = 2 \sqrt{\sin 2x}$$

Final Answer: $y = 2\sqrt{\sin 2x}$

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the method of solving differential equations by separating variables and applying initial conditions.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific algorithm or method (separation of variables and integration) to solve the differential equation.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests a standard method taught in the textbook for solving differential equations.