Problem 88 - A Complete Resource Book in Mathematics for JEE Main

Question

A differential equation of the form \(y=p x+f(p)\), where \(p=\frac{d y}{d x}\) is known as Clairaut's equation. We now find the solution of the above equation. The given equation is $$ y=p x+f(p) $$ It is solvable for \(y\). Differentiating with respect to \(x\), we get $$ \frac{d y}{d x}=p+x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x} $$ \(\Rightarrow\) $$ p=p+x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x} $$ \(\Rightarrow \quad x \frac{d p}{d x}+f^{\prime}(p) \cdot \frac{d p}{d x}=0\) Factorizing \(\frac{d p}{d x}+\left[x+f^{\prime}(p)\right]=0\) Cancelling the factor \(x+f^{\prime}(p)\) which does not involve \(\frac{d p}{d x}\), we have \(\frac{d p}{d x}=0\) Integrating, $$ p=c $$ Eliminating \(p\) between (1) and (2), the required solution of \((1)\) is $$ y=c x+f(c) $$ The solution of the equation \(p^{2} x(x-2)+p(2 y-2 x y-x+2)+y^{2}+y=0\) is (A) \((y+c x+2 c)(y-c x+1)=0\) (B) \((y-c x+2 c)(y+c x+1)=0\) (C) \((y-c x+2 c)(y-c x+1)=0\) (D) \((y-c x+2 c)(y-c x-1)=0\)

Step-by-Step Solution

Answer

The correct answer is (C): \((y - cx + 2c)(y - cx + 1) = 0\).

Step 1: Understand Clairaut's Equation

Clairaut's equation is given as \( y = px + f(p) \), where \( p = \frac{dy}{dx} \). The equation expresses \( y \) in terms of \( x \) and the slope \( p \).

Step 2: Differentiate the Equation

Differentiate both sides of the equation \( y = px + f(p) \) with respect to \( x \) to obtain \( \frac{dy}{dx} = p + x\frac{dp}{dx} + f'(p) \cdot \frac{dp}{dx} \).

Step 3: Set the Differential Equation

Using \( \frac{dy}{dx} = p \) in Clairaut's context, equate the expressions to \( p = p + x\frac{dp}{dx} + f'(p) \cdot \frac{dp}{dx} \). This simplifies to \( x\frac{dp}{dx} + f'(p) \cdot \frac{dp}{dx} = 0 \).

Step 4: Simplify and Solve

Factorize to obtain \( \frac{dp}{dx}(x + f'(p)) = 0 \). Since \( x + f'(p) \) cannot be zero for all \( p \), set \( \frac{dp}{dx} = 0 \).

Step 5: Integrate to Find General Solution

Integrate \( \frac{dp}{dx} = 0 \) to find \( p = c \), which implies that \( y = cx + f(c) \) is the general solution of Clairaut's equation.

Step 6: Solve the Given Equation

For the given equation \( p^2 x(x-2) + p(2y - 2xy - x + 2) + y^2 + y = 0 \), substitute \( p = c \) to eliminate \( p \) and reduce the equation based on the solution form \( y = cx + f(c) \).

Step 7: Simplify Options for Solutions

Substitute the solutions from the options to verify which satisfies the simplified form of the given equation. By testing, option (C), \((y - cx + 2c)(y - cx + 1) = 0\), satisfies the given differential equation.

Key Concepts

Differential Equations

Differential equations are equations that relate a function with its derivatives. In simpler terms, these equations involve variables which depend on one or more independent variables, along with their rates of change.
For example, in Clairaut's equation, we see this relationship because it contains the variable \( y \), its derivative \( p \), and additional terms like \( f(p) \).
The focus of such equations is to find the function that meets this relationship as specified in the equation.

• Understanding this definition is crucial, as it forms the backbone of many areas in physics, engineering, and mathematics.
• It explains how quantities change, predicting future behavior by modeling real-life scenarios.

Slope Function

The slope function in the context of differential equations is often represented by a derivative. In Clairaut's equation, \( p = \frac{dy}{dx} \) describes the slope or rate of change of the function \( y \) with respect to \( x \).
This tells us how steep the curve is at any given point. If \( p \) is constant, then the curve has a uniform slope.

• In Clairaut's equation, the slope \( p \) is not just any derivative—it directly appears in the equation, indicating its essential role in defining \( y \).
• Think of the slope as the geometric interpretation of the problem. It bridges the gap between the abstract equation and a graphical representation.

General Solution

The general solution of a differential equation provides a set of all possible solutions. It's like a master key opening every potential door that solves the equation under consideration.
In Clairaut's equation, after differentiating and manipulating the equation, we arrive at the solution \( y = cx + f(c) \).

• Here, \( c \) is an arbitrary constant representing an infinite number of solutions corresponding to different initial conditions or specific scenarios.
• The general solution helps us to understand all forms the equation can take, allowing us to adapt it for various applications or additional conditions.

Integration Technique

The integration technique is used to solve differential equations and involves finding the function from its derivative.
In the step-by-step solution, when we have \( \frac{dp}{dx} = 0 \), integrating gives \( p = c \).

• This step is crucial because it transitions from a rate of change back to the original function, moving from the derivative \( p \) to the function by finding \( c \).
• Integration is used widely to solve different kinds of problems where we need to construct the entire equation from smaller elements, like derivatives.