Sigma Percentile
JEE Main 2025 (January)
LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let be the solution of the differential equation and Then is equal to :

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Visualized Solution

Visualizing the Initial Value Problem

  • Given differential equation:
  • Initial condition:
  • We need to find the value of .

Analyzing the Equation's Structure

  • Notice the argument of the sine function:
  • This is a strong hint to express the entire equation in terms of .

Dividing by

  • Divide both sides by (since ):
  • Simplifies to:

The Quotient Rule Connection

  • Recall the quotient rule for differentiation:

Algebraic Manipulation for Substitution

  • Factor out from the numerator:
  • Rearranging gives:

Substituting into the Equation

  • Substitute the expression back:

Separating the Variables

  • Rearrange to separate and :

Integrating Both Sides

  • Integrate:
  • Result:
  • Simplify:

Applying the Initial Condition

  • Use to find .
  • Substitute and :

Evaluating the Constant

The Particular Solution

  • Substitute back into the equation:

Evaluating at

  • We need to find information about .
  • Substitute into the particular solution:

The Double Angle Identity

  • We need , but we have .
  • Use the identity:
  • Let , so :

Final Computation

  • Substitute into the identity:
  • This matches the correct option.

The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

The given differential equation is:
We are provided with the initial condition and our goal is to determine the value of .

The Insight

Observe the argument inside the sine function, . This suggests that the equation can be simplified by expressing it in terms of this ratio. Dividing both sides by (assuming ), we obtain:
Recall the quotient rule for the derivative of with respect to :
This allows us to rewrite the bracketed term as .

The Transformation

Substituting this expression back into our equation yields:
Rearranging the terms to separate the variables, we get:
Integrating both sides, we find:
Multiplying by , the general solution is:

The Final Act

Using the condition , we substitute and into the general solution:
Thus, , and our particular solution is .
To find when , we first note that . We apply the double angle identity :
The final result is:

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