Differential Equation | MIT 18.01SC Single Variable Calculus, Fall 2010
TLDRIn this recitation video, the professor tackles a differential equation without initial conditions, aiming to find a function y such that x times dy/dx equals x squared plus x times y squared plus 1. The solution involves the separation of variables technique, where all y-related terms are isolated on one side, and x-related terms on the other, with dx moved to the right side. The integral of 1 over y squared plus 1 is arctan(y), and the right side simplifies to x squared over 2 plus x, plus a constant. The final step involves taking the tangent of both sides to isolate y, resulting in y as the tangent of a linear function of x and a constant. The process is concluded with a suggestion to check the solution by differentiating the proposed y function.
Takeaways
- 📚 The video is a recitation focused on solving a specific differential equation.
- 🔍 The differential equation given is \( x \cdot \frac{dy}{dx} = x^2 + x \cdot y^2 + 1 \).
- 🤔 The professor encourages students to think about the problem before proceeding with the solution.
- 📐 The technique of separation of variables is used to solve the equation.
- 🔄 The terms involving \( y \) are moved to the left side, and \( dx \) is moved to the right side.
- 📉 The equation is rewritten to isolate \( \frac{dy}{y^2 + 1} \) on the left and \( x + 1 \) on the right.
- ❗ A special note is made about the potential issue when \( x = 0 \), but it's ignored for the moment.
- 📚 The left side of the equation is integrated to get \( \arctan(y) \).
- 📈 The right side is integrated to yield \( \frac{x^2}{2} + x \), with constants of integration added.
- 🔄 To isolate \( y \), the inverse function of \( \arctan(y) \) is used, which is \( \tan \).
- 📝 The final expression for \( y \) is \( \tan(\frac{x^2}{2} + x + C) \), where \( C \) is an arbitrary constant.
- 🔧 The professor suggests checking the solution by taking the derivative of the right-hand side and comparing it to the original equation.
Q & A
What is the differential equation the professor wants to solve?
-The differential equation is x * dy/dx = x^2 + x * y^2 + 1.
What method does the professor use to solve the differential equation?
-The professor uses the technique of separation of variables to solve the differential equation.
What is the first step in the separation of variables process for this equation?
-The first step is to get all the terms that involve y on the left-hand side and move the dx to the right-hand side.
How does the professor rewrite the equation after separating the variables?
-The professor rewrites the equation as dy/(y^2 + 1) = (x + 1) dx.
What is the antiderivative of 1/(y^2 + 1)?
-The antiderivative of 1/(y^2 + 1) is arctangent of y, or arctan(y).
What does the professor get after integrating both sides of the equation?
-After integrating, the professor gets arctan(y) = (x^2/2) + x + C, where C is the constant of integration.
How does the professor isolate y in the equation?
-The professor takes the tangent of both sides of the equation to isolate y, resulting in y = tan((x^2/2) + x + C).
Why can't the professor solve for the constant C?
-The professor can't solve for the constant C because no initial conditions are provided.
How can you verify that the solution is correct?
-You can verify the solution by taking the derivative of y with respect to x and checking if it satisfies the original differential equation.
What is the relationship between arctan(y) and its derivative?
-The derivative of arctan(y) with respect to y is 1/(y^2 + 1).
Outlines
📚 Differential Equation Introduction
The professor begins the recitation session by introducing a differential equation problem without initial conditions. The goal is to find a function y such that x times its derivative dy/dx equals x squared plus x times y squared plus 1. The professor encourages students to think about the problem before proceeding with the solution using the technique of separation of variables.
🔍 Separation of Variables Technique
The professor demonstrates the separation of variables technique to solve the given differential equation. He rearranges the equation to isolate terms involving y on one side and terms involving x on the other, including the differential dx. The equation is then simplified, and the professor prepares to integrate both sides, setting up the integration by considering the antiderivative of 1 over y squared plus 1 as arctangent of y.
🧑🏫 Integration and Isolation of y
After integrating both sides of the equation, the professor obtains expressions for the left and right sides involving arctangent and linear functions of x, respectively. He then isolates y by taking the tangent of both sides, leading to an expression for y in terms of the tangent function applied to a combination of x and a constant. The professor notes the absence of initial conditions, which prevents solving for the constant C in the solution.
🔧 Verification of the Solution
The professor suggests verifying the solution by taking the derivative of the proposed solution for y and comparing it to the original differential equation. This step is crucial to ensure the correctness of the solution, although the professor does not explicitly perform this verification in the script provided.
Mindmap
Keywords
💡Recitation
💡Differential Equation
💡Separation of Variables
💡Antiderivative
💡Arctangent
💡Integration
💡Tangent
💡Initial Condition
💡Constant of Integration
💡Derivative
Highlights
Introduction to solving a differential equation without initial conditions.
Equation given: x * dy/dx = x^2 + x * y^2 + 1.
Use of separation of variables technique for solving the equation.
Rearranging terms to facilitate separation of variables.
Ignoring the singularity at x = 0 for the moment.
Rewriting the equation for integration setup.
Integration of both sides of the equation.
Finding the antiderivative of 1/(y^2 + 1) as arctan(y).
Derivative of arctan(y) equals 1/(y^2 + 1).
Integration of x and x^2/2 to find the right-hand side.
Adding integration constants to both sides.
Isolating y by taking the tangent of both sides.
Final expression for y in terms of x and C.
Discussion on the absence of initial conditions affecting C.
Method to check the solution by differentiating the right-hand side.
Stopping the explanation without completing the derivative check.
Transcripts
Browse More Related Video
AP Calculus AB Crash Course Day 5 - Integration and Differential Equations
How to use implicit differentiation on an equation for dy dx
Linear versus Nonlinear Differential Equations
Transforming General Form to Standard Form of Parabola | @ProfD
Implicit Differentiation for Calculus - More Examples, #1
Worked example: Implicit differentiation | Advanced derivatives | AP Calculus AB | Khan Academy
5.0 / 5 (0 votes)
Thanks for rating: