Problem 58 We have \(\frac{d y}{d x}=1-\fra... [FREE SOLUTION]

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Calculus for JEE ADVANCED

G. Tewani

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Chapter 10: Problem 58

We have $\frac{d y}{d x}=1-\frac{1}{x^{2}}$ or $y=x+\frac{1}{x}+C$

Short Answer

Expert verified

The proposed solution satisfies the differential equation.

Step by step solution

Understand the Problem

We are given a differential equation $ \frac{dy}{dx} = 1 - \frac{1}{x^2} $ and a proposed solution $ y = x + \frac{1}{x} + C $, where $ C $ is a constant. We need to verify if the proposed solution satisfies the differential equation.

Differentiate the Proposed Solution

Differentiate the given solution $ y = x + \frac{1}{x} + C $ with respect to $ x $. The derivative of $ x $ is 1, the derivative of $ \frac{1}{x} $ is $ -\frac{1}{x^2} $, and $ C $ being a constant has a derivative of 0. Therefore, $ \frac{dy}{dx} = 1 - \frac{1}{x^2} $.

Verify the Solution

Compare the computed derivative $ \frac{dy}{dx} = 1 - \frac{1}{x^2} $ from Step 2 to the original differential equation $ \frac{dy}{dx} = 1 - \frac{1}{x^2} $. Since both expressions are identical, the proposed solution $ y = x + \frac{1}{x} + C $ satisfies the differential equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solution Verification

Verification of a solution in differential equations is a vital step to ensure correctness. When you have a proposed solution, like the function $ y = x + \frac{1}{x} + C $, to a given differential equation, such as $ \frac{dy}{dx} = 1 - \frac{1}{x^2} $, the process of verification involves checking whether the solution satisfies the equation. This is done by differentiating the solution and comparing it to the original differential expression. If both expressions match, it confirms that the solution is correct. This provides a logical closure that builds confidence in solving such equations accurately.

Constant of Integration

In calculus, the constant of integration $ C $ is crucial when working with indefinite integrals. Whenever you integrate a function, this constant appears because integration is the reverse of differentiation, which loses constant values during computation. Things to remember about the constant of integration:

It represents an infinite number of possible vertical shifts of the function along the y-axis.
Each possible value of $ C $ provides a unique solution, making it essential when solving real-world problems involving initial conditions.

In the solution $ y = x + \frac{1}{x} + C $, the constant $ C $ allows the general solution to account for different boundary conditions that may arise in practical applications.

Derivative Calculation

Calculating derivatives correctly is central to solving differential equations. In the given problem, differentiating the solution $ y = x + \frac{1}{x} + C $ involves applying simple differentiation rules.Key points in derivative calculation:

The derivative of a constant $ C $ is zero.
The derivative of $ x $ with respect to $ x $ is 1.
The derivative of $ \frac{1}{x} $ can be derived using the power rule, resulting in $-\frac{1}{x^2}$.

Using these calculations, we derive: $ \frac{dy}{dx} = 1 - \frac{1}{x^2} $, which matches the original differential equation. Practicing these calculations helps deepen understanding of derivatives and enhances problem-solving skills in calculus.

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91影视

We have \(\frac{d y}{d x}=1-\frac{1}{x^{2}}\) or \(y=x+\frac{1}{x}+C\)

Short Answer

Step by step solution

Understand the Problem

Differentiate the Proposed Solution

Verify the Solution

Key Concepts

Solution Verification

Constant of Integration

Derivative Calculation

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