Problem 84 - 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 integrating factor to make the differential equation \(\left(y^{4}+2 y\right) d x+\left(x y^{3}+2 y^{4}-4 x\right) d y=0\) exact is (A) \(\frac{1}{y}\) (B) \(\frac{1}{y^{2}}\) (C) \(\frac{1}{y^{3}}\) (D) \(\frac{1}{y^{4}}\)

Step-by-Step Solution

Answer

(C) \( \frac{1}{y^3} \)

Step 1: Identify the equation form

The given differential equation is \( (y^4 + 2y) \, dx + (xy^3 + 2y^4 - 4x) \, dy = 0 \). This matches the form \( M \, dx + N \, dy = 0 \), where \( M = y^4 + 2y \) and \( N = xy^3 + 2y^4 - 4x \).

Step 2: Compute partial derivatives

Calculate the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \).\[ \frac{\partial M}{\partial y} = \frac{d}{dy}(y^4 + 2y) = 4y^3 + 2 \]\[ \frac{\partial N}{\partial x} = \frac{d}{dx}(xy^3 + 2y^4 - 4x) = y^3 - 4 \]

Step 3: Check if it's exact using derivatives

For the equation to be exact, \( \frac{\partial M}{\partial y} \) must equal \( \frac{\partial N}{\partial x} \). However, we have:\[ 4y^3 + 2 eq y^3 - 4 \]Thus, the equation is not exact.

Step 4: Apply exactness criterion for integrating factor

We use Rule IV, where \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \) is divided by \( M \) to check if it is a function of \( y \) alone.\[ \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = \frac{y^3 - 4 - (4y^3 + 2)}{y^4 + 2y} = \frac{-3y^3 - 6}{y^4 + 2y} \]Simplify to see if it depends solely on \( y \).

Step 5: Simplify and identify dependency

Simplify \[ \frac{-3y^3 - 6}{y^4 + 2y} = \frac{-3(y^3 + 2)}{y(y^3 + 2)} = \frac{-3}{y} \].This expression is dependent solely on \( y \), meeting the criteria.

Step 6: Find the integrating factor

Since \( \frac{-3}{y} \) is a function solely of \( y \), the integrating factor based on Rule IV is:\[ \text{I.F.} = e^{\int \frac{-3}{y} \, dy} = e^{-3\ln|y|} = |y|^{-3} = \frac{1}{y^3} \]

Step 7: Match with given options

Compare the integrating factor \( \frac{1}{y^3} \) to the options provided:(A) \( \frac{1}{y} \)(B) \( \frac{1}{y^2} \)(C) \( \frac{1}{y^3} \)(D) \( \frac{1}{y^4} \). The correct option is (C) \( \frac{1}{y^3} \).

Key Concepts

Exact Differential Equations

Exact differential equations are a special class of differential equations that can be easily identified and solved using specific techniques. These equations are in the form \( M dx + N dy = 0 \). For an equation to be exact, it must satisfy the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). This means the partial derivatives of the functions involved match in a specific way.

When an equation is exact, it implies that there exists a potential function \( \Psi(x,y) \) such that \( d\Psi = M dx + N dy \). Solving an exact equation involves finding this potential function, which then allows for the determination of the solution to the differential equation.

It's important to check for exactness first, as this method can greatly simplify solving the equation. If an equation isn’t exact, it may be transformed using an integrating factor, making it possible to solve using these techniques.

Partial Derivatives

Partial derivatives play a crucial role in verifying the exactness of differential equations. They involve differentiating a multivariable function with respect to one variable while keeping the other variables constant.

In the context of differential equations, we calculate \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) to assess whether an equation is exact. These derivatives provide insight into the behavior and relationship of the functions within the equation.

Understanding how to compute and interpret partial derivatives is foundational in both mathematics and applications in science and engineering. They allow us to understand how changes in one variable affect a function, which is essential for many types of problem-solving situations.

Homogeneous Differential Equations

Homogeneous differential equations are a type of differential equation that can often be solved more simply by utilizing their inherent properties. An equation is called homogeneous if it can be expressed in terms of a single degree (or scale) across its terms.

These equations often appear as \( M(x, y) dx + N(x, y) dy = 0 \) where both \( M \) and \( N \) can be expressed as functions of the same degree in \( x \) and \( y \). Homogeneous equations may sometimes become exact after multiplying by an integrating factor.

The characteristic feature of homogeneous functions is their scaling property: if you multiply their arguments by a constant, the function scales in a predictable way. This property is particularly useful for analyzing and solving them, as it frequently paves the way for substitutions that simplify the problem into a more manageable form.