Problem 86 - 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 \(\left(\frac{d y}{d x}\right)^{2}\left(x^{2}-a^{2}\right)-2\left(\frac{d y}{d x}\right) x y+y^{2}-b^{2}=0\) is (A) \(y=c x+\sqrt{a^{2} c^{2}+b^{2}}\) (B) \(y=c x+\sqrt{b^{2} c^{2}+a^{2}}\) (C) \(y=c x-\sqrt{a^{2} c^{2}+b^{2}}\) (D) \(y=c x-\sqrt{b^{2} c^{2}+a^{2}}\)

Step-by-Step Solution

Answer

The correct choice is (A) \( y = cx + \sqrt{a^{2}c^{2} + b^{2}} \).

Step 1: Understanding Clairaut's Equation

The given equation is a Clairaut's equation of the form \( y = px + f(p) \), where \( p = \frac{dy}{dx} \). Clairaut's equation can be solved by differentiating it with respect to \( x \) and solving for \( p \).

Step 2: Differentiate Equation with Respect to x

Differentiating the equation \( y = px + f(p) \) with respect to \( x \), we have: \( \frac{d y}{d x} = p + x \frac{d p}{d x} + f^{\prime}(p) \cdot \frac{d p}{d x} \). Equating this with \( p \) gives: \( p = p + x \frac{d p}{d x} + f^{\prime}(p) \cdot \frac{d p}{d x} \).

Step 3: Simplify the Equation

Rewriting the differentiated form: \( x \frac{d p}{d x} + f^{\prime}(p) \cdot \frac{d p}{d x} = 0 \). This simplifies further implying \( \frac{d p}{d x}[x + f^{\prime}(p)] = 0 \).

Step 4: Solve for dp/dx

From \( \frac{d p}{d x} [x + f^{\prime}(p)] = 0 \), cancel the non-zero term, so \( \frac{d p}{d x} = 0 \). Thus, \( p \) is a constant, say \( c \).

Step 5: Substitute p = c into Original Equation

Substituting \( p = c \) back into the original equation \( y = px + f(p) \), we obtain \( y = cx + f(c) \). This is the general form of the solution to Clairaut's equation.

Step 6: Match the Solution to Given Options

We are given options to match our solution \( y = cx + f(c) \) to a specific form of equation \( (\frac{d y}{d x})^{2}(x^{2}-a^{2}) - 2\frac{d y}{d x}xy + y^{2} - b^{2} = 0 \). Solving that using Clairaut's result suggests \( y = cx \pm \sqrt{a^{2}c^{2} + b^{2}} \). The matching form is option (A): \( y = cx + \sqrt{a^{2}c^{2} + b^{2}} \).

Key Concepts

Differential Equations

Differential equations are equations that involve a function and its derivatives. They play a crucial role in mathematics and engineering because they describe physical phenomena such as motion, heat, and waves. Clairaut's equation is a specific type of differential equation that takes the form \( y = px + f(p) \), where \( p \) represents the derivative \( \frac{dy}{dx} \). Understanding how to handle such equations is key as they often describe real-world systems.
Clairaut's equation is intriguing because it combines variables \( x \), \( y \), and the derivative \( \frac{dy}{dx} \), offering a different perspective from standard forms. It requires a special method of solving, making it a fascinating area for exploring solution methods.

Solution Methods

Solving Clairaut's equation involves some unique steps. The first is differentiating the equation with respect to \( x \) to bring in the derivative \( \frac{dy}{dx} \). This differentiation yields another equation that typically needs simplification and manipulation to solve for \( p \).
In our specific problem, differentiating \( y = px + f(p) \) leads to \( p + x \frac{dp}{dx} + f'(p) \cdot \frac{dp}{dx} = p \). Simplifying this gives \( x \frac{dp}{dx} + f'(p) \cdot \frac{dp}{dx} = 0 \), which ultimately implies that \( \frac{dp}{dx} = 0 \), meaning that \( p \) must be a constant. This result makes the solution process more straightforward, as it reduces the problem to a more simple form.

• First, differentiate with respect to \( x \).
• Simplify the resulting equation.
• Find \( \frac{dp}{dx} \) and solve for constants.
• Substitute back to find the general solution.

When solving differential equations like Clairaut's, these methods and exercises offer invaluable insights into more advanced calculus applications.

Calculus Concepts

Calculus is deeply intertwined with the study of differential equations, as it provides the framework for dealing with changes and dynamics. In Clairaut's equation, calculus concepts such as differentiation and integration are essential.
Differentiation is the process used to obtain the rate at which one quantity changes with respect to another. In solving Clairaut's equation, the key move is differentiating the equation to manipulate and solve for the derivative \( p \).
Once we have determined that \( \frac{dp}{dx} = 0 \), integration comes into play to solve for the constant \( p \), yielding \( p = c \). This shows how integration and differentiation are complementary processes in calculus; one process helps us find rates of change, while the other tells us how much a quantity accumulates over time.

• Understand differentiation as finding rates of change.
• Use integration to determine constant solutions.
• Apply these concepts to find comprehensive solutions in differential equations.

Through exercises like the Clairaut's equation, students can see the concrete application of these calculus concepts in solving real-world problems.