Problem 17 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The solution of the differential equation \(\frac{d y}{d x}-\frac{y+3 x}{\log _{e}(y+3 x)}+3=0\) is [Sep. 04, 2020 (II)] (where \(C\) is a constant of integration.) (a) \(x-\frac{1}{2}\left(\log _{e}(y+3 x)\right)^{2}=C\) (b) \(x-\log _{e}(y+3 x)=C\) (c) \(y+3 x-\frac{1}{2}\left(\log _{e} x\right)^{2}=C\) (d) \(x-2 \log _{e}(y+3 x)=C\)

Step-by-Step Solution

Answer

The solution is option (a): \( x - \frac{1}{2}(\log_e(y+3x))^2 = C \).

Step 1: Analyze the Differential Equation

The given differential equation is \( \frac{d y}{d x} - \frac{y+3x}{\log_e(y+3x)} + 3 = 0 \). We need to examine its structure and simplify it for integration.

Step 2: Introduce Substitution

Let \( v = \log_e(y+3x) \). This implies that \( y + 3x = e^v \). Then, differentiate with respect to \(x\): \( \frac{dy}{dx} + 3 = \frac{dv}{dx}e^v \).

Step 3: Substitute into the Differential Equation

Replace \( \frac{dy}{dx} \) with \( \frac{dv}{dx}e^v - 3 \) in the original equation to get: \[ \left(\frac{dv}{dx}e^v - 3\right) - \frac{e^v}{v} + 3 = 0. \] Simplifying gives: \( \frac{dv}{dx}e^v = \frac{e^v}{v} \).

Step 4: Simplify the Equation

Cancel \(e^v\) from both sides of \( \frac{dv}{dx}e^v = \frac{e^v}{v} \), we get \( \frac{dv}{dx} = \frac{1}{v}\).

Step 5: Integrate Both Sides

Integrate \( \frac{dv}{dx} = \frac{1}{v} \) with respect to \(x\): \[ \int v \, dv = \int dx \]. The integration gives: \[ \frac{v^2}{2} = x + C, \] where \(C\) is the constant of integration.

Step 6: Back-Substitute the Variable

Recall that \( v = \log_e(y+3x) \), so we substitute back to get \( \frac{1}{2}(\log_e(y+3x))^2 = x + C \). Rearranging gives: \( x - \frac{1}{2}(\log_e(y+3x))^2 = -C \).

Step 7: Verify the Options

Check the solution \( x - \frac{1}{2}(\log_e(y+3x))^2 = -C \). This matches option (a) \( x - \frac{1}{2}(\log_e(y+3x))^2 = C \) with a constant redefined. Thus, option (a) is correct.

Key Concepts

Constant of Integration

When solving differential equations, you often come across a term called the "constant of integration." This constant represents an infinite number of potential solutions.
It is denoted by the letter \( C \). The presence of this constant accounts for the fact that indefinite integrals can have multiple solutions. Each of these solutions can be arrived at by adding or subtracting different values to \( C \).
Consider the integration step in our exercise. Here, after integration, we reached \( \frac{v^2}{2} = x + C \). This equation includes \( C \), representing any constant value that would keep the equation balanced, showing that many similar solutions exist.
In scenarios dealing with real-world problems, \( C \) can be determined if initial conditions are known. Without these, \( C \) remains an arbitrary constant highlighting the family of possible solutions.

Variable Substitution

Variable substitution is a powerful technique used to simplify differential equations, making them easier to solve. In this method, a new variable is introduced to replace a more complex expression.
This simplification helps in transforming the equation into a more recognizable form that can be managed using standard techniques.

• In the given exercise, we made a substitution where \( v = \log_e(y+3x) \).
• This substitution allowed simplification as \( y+3x \) turned to \( e^v \), simplifying the differentiation.

Substitution is particularly effective when dealing with logarithms, exponentials, or trigonometric functions as these can complicate direct calculations.
By approaching the equation with a simpler variable, the integration and subsequent back-substitution become more straightforward, allowing us to retrieve the original variables in a neat form.

Integration Techniques

Integration is a fundamental operation in calculus used to find the antiderivative or integral of functions. Different techniques are applied based on the equation's structure.
In the exercise, a standard technique known as direct integration was employed. This involved integrating \( \frac{dv}{dx} = \frac{1}{v} \) with respect to \( x \).
We integrated both sides separately:

• For the left side: \( \int v \, dv \)
• For the right side: \( \int dx \).

The antiderivative of \( v \, dv \) gives \( \frac{v^2}{2} \). Simultaneously, \( dx \) integrates to \( x + C \), yielding a linear solution.
Choosing the correct technique, like substitution or integration by parts, can dramatically simplify solving differential equations, making complex problems easier and more feasible to solve comprehensively.

Mathematical Simplification

Simplifying mathematical equations is a critical skill that involves making equations more manageable and easier to work with. This helps in finding solutions more efficiently.
In our example, simplification occurred at several crucial steps:

• First, by cancelling \( e^v \) from both sides of the equation \( \frac{dv}{dx}e^v = \frac{e^v}{v} \).
• Then, writing \( \frac{dv}{dx} = \frac{1}{v} \), which was a much simpler form to integrate.

Such simplifications are vital as they allow for clearer visualization and manipulation of the equation.
After solving the equation, rearrangements are performed to align with the given solution format, helping ensure clarity and correctness. By focusing on simplification, we can refine our approach throughout the problem-solving process, enabling more effective integrations and back-substitutions.