Problem 82 - A Complete Resource Book in Mathematics for JEE Main

Question

Passage 2 Some equations which are not exact can be made exact on multiplication by some suitable function known as an integrating factor. The equation $$ x d y-y d x=0 $$ which is not exact becomes so on multiplication by \(1 / y^{2}\), for then we $$ \frac{x}{y^{2}} d y-\frac{1}{y} d x=0 $$ which is easily seen to be exact. We can solve it either by re-arranging the terms and making them exact differential or by the method of exact equations. We now give some rules for finding integrating factors of differential equation $$ M d x+N d y=0 $$ to make it exact. I. If \(M x+N y \neq 0\) and the equation is homogeneous, then \(\frac{1}{M x+N y}\) is an I.F. II. If the equation \(M d x+N d y=0\) is not exact but is of the form \(f_{1}(x y) y d x+f_{2}(x y) x d y=0\), then \(\frac{1}{M x-N y}\) is an I.F., provided \(M x-N y \neq 0\) III. When \(\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\) is a function of \(x\) alone, say \(f(x)\), then I.F. \(=e^{\int f(x) d x}\) IV. When \(\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\) is a function of \(y\) alone, say \(f(y)\), then I.F. \(=e^{\int f(y) d y}\) The solution of the equation \((x y \sin x y+\cos x y) y d x+(x y \sin x y-\cos x y) x d y=0\) is (A) \(y \sec x y=c x\) (B) \(x \sec x y=c y\) (C) \(x \operatorname{cosec} x y=c y\) (D) None of these

Step-by-Step Solution

Answer

The solution is (B) \(x \sec xy = c y\).

Step 1: Identify the equation M dx + N dy = 0

Given the equation \((x y \sin x y+\cos x y) y d x+(x y \sin x y-\cos x y) x d y=0\), we can denote:\[M = (x y \sin x y + \cos x y)y \quad \text{and} \quad N = (x y \sin x y - \cos x y)x\]

Step 2: Check for exactness

For the equation to be exact, \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). Calculate these partial derivatives:\[\frac{\partial M}{\partial y} = y^2 \sin xy + x^2 y \cos xy + \frac{d}{dy}[y \cos xy] \quad \text{and} \quad \frac{\partial N}{\partial x} = x^2 y \cos xy.\]

Step 3: Simplify the partial derivatives

Simplify \(\frac{d}{dy}[y \cos xy]\) term:\[\frac{d}{dy}[y \cos xy] = \cos xy - xy \sin xy.\]Now the complete equation for \(\frac{\partial M}{\partial y}\) becomes:\[y^2 \sin xy + x^2 y \cos xy + \cos xy - xy \sin xy.\] Compare with \(\frac{\partial N}{\partial x} = x^2 y \cos xy\) and check for equality.

Step 4: Test homogenity to find integrating factor

Check if the provided equation is homogeneous and if \(M x + N y eq 0\). Let's see if the terms \(xy\sin xy + \cos xy\) for both \(M\) and \(N\) form a homogeneous function which they do by degree inspection, therefore the simplifying integrating factor can be \(\frac{1}{M x + N y}\).

Step 5: Apply the integrating factor for homogenity

Since the equation is homogeneous by terms and degree, the integrating factor: \[\frac{1}{M x + N y} = \frac{1}{\left((xy \sin xy + \cos xy)xy + (xy \sin xy - \cos xy)yx\right)} \Rightarrow \frac{1}{2x^2y^2 \sin xy}\] gives us the integrating factor.

Step 6: Solve the homogenized equation

Multiply both sides by the integrating factor \(\frac{1}{2x^2y^2 \sin xy}\). This simplifies the equation and allows direct integration terms to be easier noticed because terms cancel, getting:\[x \sec xy = c y\] This aligns directly with solution (B).

Step 7: Verify answer

Verify that our derived simplified form \(x \sec xy = c y\) indeed satisfies the rearranged equation upon substituting the terms back and solves the differential equation correctly.

Key Concepts

Exact Differential Equations

Exact differential equations are a class of differential equations that arise from the equality of mixed partial derivatives. When we have a differential equation in the form \( M dx + N dy = 0 \), it is considered exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). This means that the equation can be derived from a potential function \( \phi(x, y) \) where \( M = \frac{\partial \phi}{\partial x} \) and \( N = \frac{\partial \phi}{\partial y} \).
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When an equation is exact, we can integrate \( M \) with respect to \( x \) and \( N \) with respect to \( y \) to find the potential function. This potential function helps solve the equation by yielding a relationship between \( x \) and \( y \).
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In cases where the equation is not exact, like when \( \frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x} \), we can often use an integrating factor to convert it into an exact equation. Finding the integrating factor makes the equation manageable and solvable using methods suited for exact equations.

Homogeneous Equations

Homogeneous equations are another significant type of differential equation. These equations have terms in which all functions or derivatives involved have the same degree. For instance, if every term can be thought of as having the same overall degree of \( x \) and \( y \), then it's a homogeneous equation.

This concept is crucial because homogeneous differential equations can often be simplified using a change of variables or integrating factor methods. More specifically, if a differential equation \( M dx + N dy = 0 \) is homogeneous, we can find an integrating factor as \( \frac{1}{Mx + Ny} \), provided \( Mx + Ny eq 0 \).

Homogeneous equations are easier to tackle because they allow transformations or substitutions that simplify the overall problem significantly, often leading directly to a form that is solvable with standard integration techniques.

Partial Derivatives

Partial derivatives are an extension of the concept of a derivative to functions of several variables. While ordinary derivatives provide the rate of change of a function with respect to one variable, partial derivatives do this while treating all other variables as constants.

For example, in the context of the equation \( M dx + N dy = 0 \), \( \frac{\partial M}{\partial y} \) means we are finding the rate at which \( M \) changes with respect to \( y \), keeping \( x \) constant.

Partial derivatives are critical in identifying exactness of differential equations. They are also instrumental in fields like thermodynamics, where the temperature, pressure, and volume are interdependent variables. Recognizing and calculating partial derivatives enable us to test exactness or even adjust equations into an exact form using integrating factors when necessary, simplifying the path to a solution.