Problem 78 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The solution of the equation \(\frac{d y}{d x}+x=x e^{(n-1) y}\) is (A) \(\frac{1}{(n-1)} \log \left(\frac{e^{(n-1) y}-1}{e^{(n-1) y}}\right)=\frac{x^{2}}{2}+c\) (B) \(e^{(n-1) y}=c e^{(n-1) y+(n-1)^{\frac{x^{2}}{2}}}+1\) (C) \(\log \left(\frac{e^{(n-1) y}-1}{(n-1) e^{(n-1) y}}\right)=n^{2}+c\) (D) \(e^{(n-1) y}=c e^{(n-1) \frac{x^{x}}{2}+x}+1\)

Step-by-Step Solution

Answer

The solution is (A): \( \frac{1}{(n-1)} \log \left( \frac{e^{(n-1) y}-1}{e^{(n-1) y}} \right) = \frac{x^2}{2} + c \).

Step 1: Identify the Type of Differential Equation

The given equation \( \frac{d y}{d x}+x=x e^{(n-1) y} \) is a first-order differential equation. Our goal is to find the general solution for \( y(x) \).

Step 2: Re-arrange Terms

Start by rearranging the given equation: \( \frac{d y}{d x} + x = x e^{(n-1) y} \). Move all the terms involving \( y \) to one side: \( \frac{d y}{d x} = x (e^{(n-1) y} - 1) \).

Step 3: Separate Variables

Separate the variables by moving all terms involving \( y \) to one side and \( x \) to the other side: \( \frac{1}{e^{(n-1) y} - 1} \frac{d y}{d x} = x \).

Step 4: Integrate Both Sides

Integrate both sides of the separated equation: \( \int \frac{1}{e^{(n-1) y} - 1} \, dy = \int x \, dx \).

Step 5: Solve the Integrals

Calculate the integrals: The left-hand side integral involves logarithmic functions, leading to \( \frac{1}{(n-1)} \log \left| e^{(n-1) y} - 1 \right| \). The right-hand side integral becomes \( \frac{x^2}{2} + C \).

Step 6: Combine Results into the General Solution

Set the integrated expressions equal to each other to form the general solution: \( \frac{1}{(n-1)} \log \left| e^{(n-1) y} - 1 \right| = \frac{x^2}{2} + C \). Simplify and combine terms where possible.

Key Concepts

First-Order Differential Equation

A first-order differential equation involves derivatives of the first degree with respect to one variable. In our context, we're dealing with an equation of the form \( \frac{dy}{dx} + p(x) = q(y) \). These equations are crucial in modeling problems where the rate of change depends on the function itself and another independent variable, usually time or space.
Understanding the nature of first-order differential equations helps us to better solve them, since knowing that they only involve first derivatives allows us to use specific strategies that might not work on higher-order equations. With our specific example, the equation \( \frac{d y}{d x}+x=x e^{(n-1) y} \) falls into this category.
This equation requires us to find a function \( y(x) \) such that this differential equation holds true for all \( x \) values in our domain. This is called finding the general solution which provides a family of functions fitting the described rate of change.

Variable Separation

The method of variable separation is a powerful tool to solve first-order differential equations that can be expressed as a product of a function of \( x \) and a function of \( y \). This technique involves rearranging the equation so that all terms involving \( y \) are on one side of the equation and all terms involving \( x \) are on the other side.

• The key is ensuring each side of the equation is only in terms of one variable, allowing us to integrate each side separately.
• In our equation \( \frac{d y}{d x} = x (e^{(n-1) y} - 1) \), we achieve separation by rearranging it into \( \frac{1}{e^{(n-1) y} - 1} \frac{d y}{d x} = x \).

The integration then becomes simpler because we've simplified our equation to two separate integrals, one for each variable. Understanding how to separate variables is key to solving many classes of differential equations, making it a cornerstone technique taught early in differential calculus courses.

General Solution

The general solution to a differential equation is a formula representing all possible solutions for the equation. It typically involves a constant \( C \), representing an infinite number of solutions corresponding to different initial conditions.
For our exercise, after integrating both sides of the equation \( \int \frac{1}{e^{(n-1) y} - 1} \, dy = \int x \, dx \), we find that:

• The left-hand side integrates into a logarithmic expression, specifically \( \frac{1}{(n-1)} \log \left| e^{(n-1) y} - 1 \right| \).
• The right-hand side integrates to \( \frac{x^2}{2} + C \).

Combining these results provides the general solution \( \frac{1}{(n-1)} \log \left| e^{(n-1) y} - 1 \right| = \frac{x^2}{2} + C \). This solution can represent multiple scenarios in which the differential equation can be solved, depending on the value of \( C \). Understanding the concept of a general solution is vital because it helps us see the broad applicability of solutions in real-world problems requiring flexibility based on initial conditions or other influences.