Problem 85 - 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 \((y-p x)(p-1)=p\) is (A) \(y=c x+\frac{c}{c-1}\) (B) \(y=c x-\frac{c}{c-1}\) (C) \(x=c y+\frac{c}{c-1}\) (D) None of these

Step-by-Step Solution

Answer

(A) \(y = cx + \frac{c}{c-1}\)

Step 1: Understand Clairaut's Equation

The given equation is a Clairaut's equation in the form \( y = px + f(p) \), where \( p = \frac{dy}{dx} \). This type of equation relates \( y \), \( x \), and \( p \).

Step 2: Differentiate with respect to x

Differentiate the given equation with respect to \( x \): \[ \frac{d}{dx}(y) = \frac{d}{dx}(px + f(p)) \]which results in \[ \frac{dy}{dx} = p + x\frac{dp}{dx} + f'(p) \cdot \frac{dp}{dx} \].

Step 3: Simplify the differentiated equation

Set \( \frac{dy}{dx} = p \) from the equation since \( \frac{dy}{dx} = p \):\[ p = p + x\frac{dp}{dx} + f'(p) \cdot \frac{dp}{dx} \].This gives:\[ x \frac{dp}{dx} + f'(p) \cdot \frac{dp}{dx} = 0 \].

Step 4: Solve for dp/dx

Factor out \( \frac{dp}{dx} \) from the simplified equation:\[ \frac{dp}{dx} (x + f'(p)) = 0 \].This implies that either \( \frac{dp}{dx} = 0 \) or \( x + f'(p) = 0 \).

Step 5: Consider case \( \frac{dp}{dx} = 0 \)

If \( \frac{dp}{dx} = 0 \), it implies \( p \) is constant, i.e., \( p = c \).

Step 6: Substitute the constant back into the original equation

Substitute \( p = c \) into the original equation to find \( y \):\[ y = cx + f(c) \].

Step 7: Solve for particular solution given in problem

The problem gives a specific equation \((y-px)(p-1)=p\) that we need to solve using \( y=cx+f(c) \). Substitute \( p \) and \( y \) back into the equation, simplifying it, and compare with given options.

Step 8: Compare and find the correct option

After substituting the expressions and simplifying, you obtain \( y = cx + \frac{c}{c-1} \), which matches choice (A).

Key Concepts

differential equations

Differential equations are mathematical equations that relate a specific function with its derivatives. They're crucial in modeling various phenomena, from physical systems to biological processes. At their core, differential equations offer insights into how a quantity changes over time or space. For instance, how the velocity of a moving object changes over time can be described by a differential equation.

Solving a differential equation typically involves finding a function that satisfies the given relationship. Some key types include ordinary differential equations (ODEs) and partial differential equations (PDEs). ODEs deal with functions of a single variable and their derivatives, whereas PDEs involve multiple variables. These equations are used across numerous fields such as physics, engineering, economics, and more.

Understanding the basics of differential equations is fundamental, as they form the backbone of understanding dynamic systems. They help predict future behavior based on initial conditions, making them invaluable in both theoretical studies and practical applications.

solution of Clairaut's equation

Clairaut's equation is a fascinating type of differential equation, with the form \( y = px + f(p) \), where \( p = \frac{dy}{dx} \). This unique structure requires a special approach to find solutions. The method primarily involves differentiating the equation with respect to \( x \) and performing algebraic manipulations.

The core steps begin with differentiating the equation, which sets the groundwork for simplifying and solving it. The next step involves setting \( \frac{dy}{dx} = p \), which simplifies the differentiated equation to either \( x \frac{dp}{dx} + f'(p) \cdot \frac{dp}{dx} = 0 \) or finding \( \frac{dp}{dx} \).

• Two cases typically arise: setting \( \frac{dp}{dx} = 0 \) assumes \( p \) is constant. This leads to \( p = c \) where \( c \) is a constant, and subsequently provides a general solution \( y = cx + f(c) \).
• Another approach may involve solving for \( x + f'(p) = 0 \), potentially providing different solutions or conditions depending on the form of \( f(p) \).

Clairaut's equations provide a structured path to finding explicit solutions, particularly for specific cases. As seen in exercises, they can lead to comparing solutions with predefined forms to ascertain correctness.

integration in calculus

Integration is a fundamental concept in calculus that deals with finding accumulation or the total of quantities. Whether calculating the area under a curve, the total distance traveled, or the net change in a process, integration plays a pivotal role.

In the context of differential equations, integration is often the step following differentiation. For example, once you've found \( \frac{dp}{dx} = 0 \), integration allows you to find constants or other terms in solutions. It transforms a differential equation from the form "rate of change" back into a "function" form.

There are different techniques for integration, such as:

• Indefinite Integration: Finding a general function, often resulting in a function plus a constant \( C \).
• Definite Integration: Calculating an exact numeric value, often used for physical problems determining the total change over an interval.
• Partial Integration: A method used when dealing with products of functions.

Through integration, we reverse the process of differentiation, allowing us to reconstruct original quantities from rates. It's a brilliant technique that sees everyday application in science, engineering, and economics, helping us understand and manipulate the world mathematically.