First Order Linear Differential Equations

The Organic Chemistry Tutor

24 Mar 201822:28

EducationalLearning

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TLDRThis video script provides a comprehensive guide on solving first-order linear differential equations. It outlines the necessary steps: writing the equation in standard form, identifying functions p(x) and q(x), determining the integrating factor, and applying it to find the general solution. The script includes detailed examples with step-by-step explanations, demonstrating how to simplify and integrate to arrive at the final solutions. It also emphasizes the importance of verifying solutions by plugging them back into the original equation.

Takeaways

📝 The process begins by writing the first-order linear differential equation in standard form: y' + p(x)y = q(x).
🔍 Identify the functions p(x) and q(x) from the standard form of the equation.
🧮 Determine the integrating factor, which is e^(∫p(x)dx).
📈 The general solution is given by 1/integrated factor * ∫[i(x)*q(x)dx] + constant of integration.
🌟 For the given example y' + 2y = 2e^x, p(x) = 2 and q(x) = 2e^x, and the integrating factor is e^(∫2dx) = e^(2x).
📚 Solving the first example yields y = (2/3)e^(x) + (c/e^(2x)).
📝 For the second example x*y' + 4y = 2x^3, after putting it in standard form and finding the integrating factor, the solution is y = (2/7)x^3 + (c/x^4).
🔄 To check the solution, plug it back into the original equation to ensure both sides are equal.
📝 For the equation (x-2)y' + y = x^2 - 4, after transforming and finding the integrating factor, the solution is y = (x^3/3) + c*(x-2).
📝 In the case of a non-homogeneous equation like 4y - 3x dx + 5x dy = 0, first transform it into a homogeneous form by dividing by dx and rearranging terms.
📝 The final example, dy/dx = (x^2 - y)/x, when put into standard form and solved, gives y = (1/3)x^3 + c*(x^-1).

Q & A

What is the standard form of a first order linear differential equation?
-The standard form of a first order linear differential equation is y' + p(x)y = q(x).
How do you identify the functions p(x) and q(x) in the differential equation?
-In the standard form y' + p(x)y = q(x), p(x) is the coefficient of y and q(x) is the function on the right side of the equation.
What is the integrating factor in the context of solving first order linear differential equations?
-The integrating factor is given by e^(∫p(x)dx), which is used to simplify the process of solving the differential equation.
How do you write the general solution of a first order linear differential equation?
-The general solution is written as (1/integrating factor) * (∫(integrating factor * q(x))dx) + C, where C is the constant of integration.
What is the significance of adding the constant of integration at the end of the solution process?
-The constant of integration is added at the end to account for the arbitrary nature of the solution, making it a general solution applicable to all initial conditions.
In the given example with y' + 2y = 2e^x, what are the values of p(x) and q(x)?
-In this example, p(x) is 2 and q(x) is 2e^x.
How does the integrating factor help in solving the differential equation?
-The integrating factor simplifies the equation by making it possible to remove the y-term, allowing for the direct integration of the remaining equation to find the solution.
What is the final solution to the differential equation y' + 4/x*y = 2x^3?
-The final solution is y = (2/7)x^3 + C/x^4, where C is the constant of integration.
How can you verify the correctness of the solution to a differential equation?
-You can verify the solution by plugging it back into the original equation and checking if both sides are equal.
What is the method to solve a differential equation when it's not in the standard form?
-First, you need to manipulate the equation to get it into the standard form y' + p(x)y = q(x). This may involve dividing by a common factor or rearranging terms.
In the case of the differential equation 4y - 3x * dx + 5x * dy = 0, how do you proceed to solve it?
-You need to rearrange the equation to get y' in terms of dx and then divide all terms by a common factor to get it into the standard form before proceeding with the solution.

Outlines

00:00

📚 Introduction to Solving First Order Linear Differential Equations

This paragraph introduces the process of solving first order linear differential equations. It emphasizes the importance of writing the equation in standard form (y' + p(x)y = q(x)), identifying the functions p(x) and q(x), and determining the integrating factor (e^(∫p(x)dx)). The paragraph then outlines the steps to find the general solution of the differential equation, which involves integrating the function q(x) and adding the constant of integration at the end. An example is provided to illustrate the process, where the equation y' + 2y = 2e^x is solved, and the integrating factor e^(2x) is used to find the general solution.

05:03

🔍 Working with Non-Standard Form Equations

This paragraph discusses how to handle differential equations that are not in standard form. It begins by transforming the equation x^2y' + 4y = 2x^3 into standard form by dividing through by x. The paragraph then identifies p(x) = 4/x and q(x) = 2x^2, and proceeds to find the integrating factor, which is x^4. The solution process is detailed, showing the integration steps and the final answer y = (2/7)x^3 + (c/x^4). The correctness of the solution is verified by plugging it back into the original equation.

10:04

🧩 Solving a Differential Equation with Factorization

This paragraph tackles a differential equation x - 2y' + y = x^2 - 4, which is solved by first putting it in standard form and factorizing the right-hand side. The standard form is derived as y' + (1/(x-2))y = (x+2)(x-2). The integrating factor is found to be x-2, and the solution is constructed using the integrating factor method. The final solution is given as y = (x^3/3) + (c/(x-2)).

15:06

📈 Converting Non-Standard Forms to Standard Form

The paragraph addresses the method of converting non-standard forms of differential equations to standard form. An example is given where the equation 4y - 3x(dy/dx) + 5x = 0 is transformed into standard form by dividing all terms by dx and rearranging. The standard form is 4y + (4/5x)y' = 3x/5. The integrating factor is calculated as x^(4/5), and the general solution is derived. The final answer is y = (3/5)x + (c/x^(4/5)).

20:07

🎓 Final Example and Summary of the Process

The final paragraph provides another example of solving a first order linear differential equation, this time with the equation dy/dx = (x^2 - y)/x. The equation is rearranged and simplified to x^2y' + y = x^2, and then put into standard form y' + (1/x)y = x. The integrating factor x is found, and the solution process is detailed, leading to the final answer y = (1/3)x^3 + c/x. The paragraph concludes by encouraging further practice with more complex problems to solidify the understanding of solving first order linear differential equations.

Mindmap

Keywords

💡First Order Linear Differential Equations

First Order Linear Differential Equations are equations that involve an unknown function and its first derivative, both in the form of a linear relationship. These equations are foundational in mathematical analysis and are used to model various phenomena in physics, engineering, and other sciences. In the video, the process of solving such equations is the main focus, with detailed steps provided for finding solutions.

💡Standard Form

In the context of differential equations, the standard form refers to the conventional way of writing an equation that aligns with a specific pattern, making it easier to solve. For first order linear differential equations, the standard form is y' + p(x)y = q(x), where p(x) and q(x) need to be identified to proceed with the solution process.

💡Integrating Factor

The integrating factor is a crucial component in the solution of first order linear differential equations. It is a function used to simplify the equation and facilitate the integration process. The integrating factor, denoted as μ(x), is determined by the integral of the function p(x), and it is used to transform the given equation into a form that can be easily integrated.

💡General Solution

The general solution of a differential equation is a formula that represents all possible solutions to the equation. It typically includes an arbitrary constant, which can be specified for particular initial conditions. In the video, the general solution is derived by integrating the product of the integrating factor and the function q(x), plus an integration constant.

💡Integration

Integration is a fundamental process in calculus that involves finding the function whose derivative is a given function. In the context of differential equations, integration is used to find the original function from its derivative. The video demonstrates the integration process as part of solving first order linear differential equations.

💡Constants of Integration

Constants of integration are added to the antiderivative of a function to account for the infinite number of possible solutions that can exist for a given differential equation. They represent the arbitrary values that can be assigned to the solution based on specific initial conditions or boundary values.

💡Antiderivatives

An antiderivative, also known as an indefinite integral, is a function whose derivative is a given function. In solving differential equations, finding the antiderivative of a function is a crucial step in integrating the equation to find its solution.

💡Factoring

Factoring is the process of breaking down a polynomial into a product of other polynomials or factors. It is a useful technique in algebra and is applied in solving certain types of differential equations, especially when they can be simplified through the factoring of terms.

💡Substitution

Substitution is a mathematical method used to replace one expression with another, often to simplify an equation or to make a problem more manageable. In the context of differential equations, substitution can be used to transform the equation into a more recognizable form or to isolate variables.

💡Power Rule

The power rule is a fundamental rule in calculus that states that the derivative of a function y = x^n, where n is a constant, is y' = n*x^(n-1). This rule is essential for finding derivatives and, by extension, for solving differential equations that involve polynomial functions.

Highlights

The video discusses solving first order linear differential equations.

The standard form of a first order linear differential equation is y prime plus p(x) times y equals q(x).

Identify functions p(x) and q(x) as the first step in solving the equation.

The integrating factor is determined by e raised to the integral of p(x) dx.

The general solution involves 1 divided by the integrating factor times the integral of i(x) times q(x) dx plus a constant of integration.

An example is given where y prime plus 2y equals 2e to the x.

The integrating factor for the example is e to the 2x.

The solution to the example is y equals 2/3 times e to the x plus c times e to the negative 2x.

Another problem is solved where x times y prime plus 4y equals 2x cubed.

The integrating factor for this problem is x to the fourth.

The solution is y equals 2/7 times x cubed plus c over x to the fourth.

The video explains how to check the solution by plugging it back into the original equation.

A third example involves x minus 2 times y prime plus y equals x squared minus 4.

The integrating factor for this example is x minus 2.

The solution to this example is y equals (x squared minus 4) divided by (x minus 2).

A fourth problem is solved with a non-standard form, 4y minus 3x times dx plus 5x dy equals 0.

The integrating factor for this non-standard form is x to the 4/5 power.

The solution is y equals x over 3 plus c times x to the negative 4/5.

The last problem solved involves dy/dx equals (x squared minus y) divided by x.

The integrating factor for this problem is x.

The solution is y equals 1/3 times x cubed plus c times x to the negative 1.

The video concludes by encouraging more practice with textbook problems.

Transcripts

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