Problem 106 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(x \frac{d y}{d x}=y(\log y-\log x+1)\), then the solution of the equation is (A) \(y \log \left(\frac{x}{y}\right)=c x\) (B) \(x \log \left(\frac{y}{x}\right)=c y\) (C) \(\log \left(\frac{y}{x}\right)=c x\) (D) \(\log \left(\frac{x}{y}\right)=c y\)

Step-by-Step Solution

Answer

The solution is option (C): \( \log \left(\frac{y}{x}\right) = cx \).

Step 1: Identify the type of differential equation

We are given the differential equation: \( x \frac{dy}{dx} = y( \log y - \log x + 1 ) \). The form \( x \frac{dy}{dx} = y( \log y - \log x + 1 ) \) suggests a need for separation of variables or possibly substitution.

Step 2: Rearrange the equation

Reorganize the terms to express the differential equation in a solvable format: \( x \frac{dy}{dx} = y( \log \frac{y}{x} + 1 ) \). From here, introduce substitution for easier manipulation.

Step 3: Use substitution to simplify

Let \( v = \log \frac{y}{x} \). Then, \( y = xe^v \) and differentiate both sides with respect to \( x \): \( \frac{dy}{dx} = e^v + xe^v \frac{dv}{dx} \). Substitute back into the original differential equation.

Step 4: Derive separable differential equation

Substitute \( \frac{dy}{dx} = e^v + xe^v \frac{dv}{dx} \) into the equation, simplify to find a differential in terms of \( v \) and \( x \): \( v = -1 \). This suggests \( \log \frac{y}{x} = c \), where \( c \) is a constant.

Step 5: Solve for y in terms of x

The result \( \log \frac{y}{x} = c \) implies \( \frac{y}{x} = e^c \). Therefore, \( y = cx \), indicating a direct relationship between \( y \) and \( x \).

Step 6: Match solution to given options

The solution \( \log \frac{y}{x} = cx \) corresponds to option (C) \( \log \left(\frac{y}{x}\right) = cx \).

Key Concepts

Separation of Variables

One of the fundamental techniques for solving differential equations is Separation of Variables. This method is often used when you can express both sides of the equation in terms of different variables. The basic goal is to isolate each variable on different sides of the equation. This approach is typically applicable to first-order differential equations.

When using Separation of Variables, you first identify if the equation can be rearranged so that each variable and its differential are on opposite sides. For example, if you have a differential equation like \( x \frac{dy}{dx} = y(\log y - \log x + 1) \), you would aim to express it such that:

• All terms involving \(y\) and \(dy\) are on one side
• All terms involving \(x\) and \(dx\) are on the other side

Finally, you integrate both sides with respect to their respective variables. The integration will likely result in one or more constant terms, and these constants are crucial for deriving the general solution. Some equations might also require a substitution before they become separable, as seen in our original exercise.

Substitution Method

The Substitution Method is a clever approach to simplifying a complex differential equation. By substituting a new variable, you transform the equation into a simpler form, making it easier to solve.

In many cases, substitution involves identifying a pattern or form within the equation that hints at a transformation. For instance, in the solution process of our original exercise, we used the substitution \( v = \log \frac{y}{x} \). This transformation simplifies multiplication or division processes because logarithmic identities convert them into addition or subtraction, respectively.

When using the substitution, differentiate the function you've substituted to find terms like \( \frac{dv}{dx} \), which can then be replaced back into the equation. After substituting and simplifying, you aim to revert to the original variable terms to solve for the desired function. This method is particularly effective when dealing with differential equations involving exponential or logarithmic components.

Solving Logarithmic Equations

Logarithmic equations, characterized by the presence of variables within a logarithm, require a specific solving strategy. Often, they're solved through the use of properties of logarithms to simplify and rewrite expressions.

In differential equations featuring logarithmic terms, identifying and utilizing these properties is vital. For the original exercise, recognizing that we had \( \log \frac{y}{x} \) was key. Such expressions in differential equations can often be simplified or re-expressed to reveal a clearer path to the solution.

• Basic property: \( \log a - \log b = \log \frac{a}{b} \)
• Antilog: Solving equations that involve \( \log y = c \) often requires exponentiating both sides to solve for \( y \).

Once the logarithms are addressed, the resulting equation is typically in a simpler algebraic form which can be solved for the unknowns easily. Finally, interpreting the results is crucial to ensuring they match any given conditions or constraints of the original problem.