Problem 51 - A Complete Resource Book in Mathematics for JEE Main

Question

Solution of equation \(\frac{d y}{d x}=\frac{y \frac{d(\phi(x))}{d x}-y^{2}}{\phi(x)}\) is (A) \(y=\frac{\phi(x)+c}{x}\) (B) \(y=\frac{\phi(x)}{x}+c\) (C) \(y=\frac{\phi(x)}{x+c}\) (D) \(y=\phi(x)+x+c\)

Step-by-Step Solution

Answer

(C) \(y=\frac{\phi(x)}{x+c}\).

Step 1: Identify the Differential Equation

The given differential equation is \( \frac{d y}{d x} = \frac{y \frac{d(\phi(x))}{d x} - y^{2}}{\phi(x)} \). This can be expressed in a separable form or checked for an integrating factor.

Step 2: Transform the Equation

Observe that the differential equation can be simplified and rearranged to separate the variables or to fit a standard form.

Step 3: Recognize as Bernoulli's Equation

Rewrite the equation in the form of a Bernoulli differential equation. A Bernoulli equation has the form \( \frac{d y}{d x} + P(x)y = Q(x)y^n \), where \( n eq 0, 1 \). This equation can be expressed as \( \frac{d y}{d x} = \frac{y \phi'(x)}{\phi(x)} - \frac{y^{2}}{\phi(x)} \).

Step 4: Solve Using Bernoulli's Method

Convert to a linear differential equation by substituting \( v = y^{1-n} = y^{-1} \), giving \( \frac{d v}{d x} = -v \frac{\phi'(x)}{\phi(x)} + \frac{1}{\phi(x)} \). This is now a first-order linear differential equation, which you can solve using an integrating factor.

Step 5: Determine Integrating Factor

Find the integrating factor \( \mu(x) = e^{\int -\frac{\phi'(x)}{\phi(x)} \, dx} = \frac{1}{\phi(x)} \), which simplifies the differential equation further.

Step 6: Integrate and Solve for v

Integrate both sides with respect to \( x \): \( \frac{d}{dx}(v \cdot \frac{1}{\phi(x)}) = \frac{1}{\phi(x)^{2}} \) leading to \( v \cdot \frac{1}{\phi(x)} = \frac{1}{x + c} \).

Step 7: Solve for y

Re-substitute \( v = y^{-1} = \phi(x) \frac{1}{x + c} \). Therefore, \( y = \frac{\phi(x)}{x + c} \).

Step 8: Conclusion: Choose the Correct Answer

After solving for \( y \), the correct solution matches option (C): \( y = \frac{\phi(x)}{x + c} \).

Key Concepts

First-order linear differential equation

Differential equations can often look a bit daunting, but breaking them down helps a lot. A first-order linear differential equation takes the form \( \frac{dy}{dx} + P(x)y = Q(x) \). Here, \( P(x) \) and \( Q(x) \) are functions of \( x \), while \( y \) is the function we are trying to determine. The equation is "first-order" because the highest derivative, \( \frac{dy}{dx} \), is first-order. It's called "linear" because each term involving \( y \), including those multiplied by its derivatives, is to the first power.

In step 4 of our exercise, we transformed a Bernoulli differential equation into a first-order linear differential equation by using a substitution method. This allowed the equation to be solved systematically. Understanding the structure of these equations helps students apply solving methods like separation of variables or using an integrating factor.

This form is particularly useful because of the techniques available to solve it, making it a cornerstone in learning about differential equations.

Integrating factor

The integrating factor is a function used to simplify differential equations, making them easier to solve. It is especially useful for solving first-order linear differential equations. The goal of using an integrating factor is to transform the differential equation into a form where both sides can be straightforwardly integrated.

In our step-by-step solution, the integrating factor \( \mu(x) \) was found to be \( e^{\int -\frac{\phi'(x)}{\phi(x)} \, dx} \), which simplifies to \( \frac{1}{\phi(x)} \). By multiplying through the differential equation by this integrating factor, we convert the left-hand side into a derivative of a product of functions.

This method helps students understand that even complex-looking equations can have elegant solutions with the right transformations, and it's an essential technique in the toolkit for solving differential equations.

Separation of variables

Separating variables is a fundamental method for solving differential equations, particularly useful when an equation can be expressed in the form where all terms of one variable are on one side of the equation, and all terms of the other are on the opposite. The process involves manipulating the equation to achieve this separation and then integrating both sides independently.

In our original exercise, separation of variables can be identified as a possible approach after simplifying and transforming the equation. By recognizing how to isolate terms related to \( y \) on one side and \( x \) on the other, it shows flexibility in solving differential equations when applicable.

Learning to use separation of variables is useful not only for specific equations but also for developing an intuitive feel for how differential equations behave and how they can be managed. This method underscores the importance of understanding the structure of an equation before deciding on the solving technique.