Problem 54 - Calculus for JEE ADVANCED

Question

\(\frac{d x}{d y}=\frac{x+2 y^{3}}{y}\)

Step-by-Step Solution

Answer

The solution is \(x = ky \cdot e^{\frac{2}{3}y^3}\), where \(k\) is a constant.

Step 1: Identify the Formula to Use

Recognize that we need to separate variables to solve the differential equation \(\frac{dx}{dy} = \frac{x + 2y^3}{y}\). Our goal is to express the equation in a form where all \(x\) terms are on one side and all \(y\) terms are on the other.

Step 2: Separate Variables

To separate variables, rewrite the equation as \(\frac{dx}{dy} = \frac{x}{y} + 2y^2\). Now, rearrange it to position \(x\) and \(y\) terms on opposite sides: \(\frac{dx}{x} = \frac{1}{y} \, dy + 2y^2 \, dy\).

Step 3: Integrate Both Sides

Integrate both sides with respect to their respective variables. For the left side, integrate \(\int \frac{1}{x} \, dx = \ln |x| + C_1\). For the right side, integrate \(\int \frac{1}{y} \, dy + \int 2y^2 \, dy = \ln |y| + \frac{2}{3}y^3 + C_2\).

Step 4: Combine the Integrals

Combine the results from the previous integrations: \(\ln |x| = \ln |y| + \frac{2}{3}y^3 + C\) where \(C = C_2 - C_1\).

Step 5: Exponentiate to Solve for x

Exponentiate both sides to solve for \(x\): \(|x| = e^{\ln|y| + \frac{2}{3}y^3 + C} = e^C \cdot |y| \cdot e^{\frac{2}{3}y^3}\). Let \(e^C = k\), where \(k\) is a constant, yielding \(x = ky \cdot e^{\frac{2}{3}y^3}\).

Step 6: Finalize the Solution

The solution to the differential equation \(\frac{dx}{dy} = \frac{x + 2y^3}{y}\) is \(x = ky \cdot e^{\frac{2}{3}y^3}\), where \(k\) is an arbitrary constant.

Key Concepts

Variable Separation

Variable separation is a method used to solve certain types of differential equations. The main idea is to rearrange the equation so that each variable appears on a different side of the equation. In our exercise, the differential equation \( \frac{dx}{dy} = \frac{x + 2y^3}{y} \) is initially mixed with both \( x \) and \( y \) terms together. This can make it harder to directly solve or integrate.To separate the variables, we rewrite the right side as a sum, \( \frac{x}{y} + 2y^2 \). Then, by rewriting the equation as \( \frac{1}{x} \, dx = \left( \frac{1}{y} + 2y^2 \right) \, dy \), we place all terms involving \( x \) on one side and all terms involving \( y \) on the other side. This separation sets the stage for integration, allowing us to address each variable independently.

Integration

Integration is a fundamental concept in calculus used to find the accumulated change and is critical to solving separated differential equations. After separating the variables in our equation, we integrate both sides independently. For the left side \( \int \frac{1}{x} \, dx \), the integration results in the natural logarithm of the absolute value of \( x \), or \( \ln |x| \).On the right side, we integrate \( \int (\frac{1}{y} + 2y^2) \, dy \). This breaks down into two simpler integrals:

• \( \int \frac{1}{y} \, dy = \ln |y| \)
• \( \int 2y^2 \, dy = \frac{2}{3} y^3 \)

After integration, we can add a generic constant to represent the indefinite nature of our calculations, denoted \( C \). This constant accounts for the initial conditions of the problem.

Exponentiation

Exponentiation is the process of raising a quantity to a power and plays a crucial role especially when reversing logarithms in the context of solving differential equations. In our case, after obtaining a logarithmic expression from integration, we need to solve for \( x \):\[ \ln|x| = \ln|y| + \frac{2}{3}y^3 + C \]The direct approach is to "exponentiate" both sides, which involves applying the exponential function \( e \) leading to:\[ |x| = e^{\ln|y| + \frac{2}{3}y^3 + C} = e^C \cdot |y| \cdot e^{\frac{2}{3}y^3} \]Here, \( e^C \) is simplified to a new constant \( k \), to make the expression clear and tidy, as \( x = ky \cdot e^{\frac{2}{3}y^3} \). This exponentiation essentially "undoes" the logarithm and results in a simpler, manageable expression.

General Solution

In the context of differential equations, a general solution includes all possible particular solutions of the differential equation. It generally consists of the expression that encompasses an arbitrary constant, representing the family of curves satisfying the differential equation.For our differential equation, the solution we derived is:\[ x = ky \cdot e^{\frac{2}{3}y^3} \]This equation represents the general solution, with \( k \) serving as an arbitrary constant. The constant \( k \) is crucial as it allows the equation to adapt to satisfy various initial conditions or specific scenarios given in different problems. Each distinct \( k \) essentially corresponds to a unique curve on a graph, forming a comprehensive set of solutions.

Differential Calculus

Differential calculus focuses on the concept of how things change, which we describe using derivatives. It is essential when dealing with differential equations like the one in our exercise. The derivative \( \frac{dx}{dy} \) is a fundamental part of the equation.Differential calculus helps us understand the rate of change. In this case, it helps describe how \( x \) changes with respect to changes in \( y \), initially provided by \( \frac{dx}{dy} = \frac{x + 2y^3}{y} \).By utilizing this form of calculus, we are able to transform the problem through steps of separating variables, integrating, exponentiating, and eventually finding solutions that detail how one variable affects the other across a given domain. Understanding these changes helps in linking different parts of a system and analyzing how one part affects another over time.