4 Solutions of FODE with Homogeneous Coefficients

Name: 4 Solutions of FODE with Homogeneous Coefficients
Uploaded: 2020-09-24T16:52:50.000Z
Duration: 1 h 4 min 12 s

Introduction to First Order Differential Equations with Homogeneous Coefficients

In this section, the speaker introduces the objective of solving first-order differential equations with homogeneous coefficients. The general form of a first-order differential equation is also discussed.

General Form of First-Order Differential Equation

A first-order differential equation has the standard form: a(x)y dx + b(x)y dy = 0.

If it is not possible to separate the variables, we can try to make it purely a function of x or y by dividing by a certain function.

If the coefficients are of the same degree in x and y, then they are said to be homogeneous.

Homogeneous Coefficients

When a and b are of the same degree in x and y, then they are said to be homogeneous.

To solve such differential equations, we can use substitution by equating one variable in terms of another.

Illustration on Homogeneous Differential Equations

In this section, an illustration is given on how to solve homogeneous differential equations using substitution.

Example Problem

Given: 2t^3 ds - (t^3 s^2 - 2s^3) dt = 0

The coefficients are homogeneous since both terms have degree 3.

Substitution Method

We can use substitution by setting s = qt where q can be any variable.

Then we get ds = qdt + tdq using derivative of a product rule.

By substituting s with qt and ds with qdt + tdq into the original equation, we get an equation that can be simplified and solved for q(t).

Solution

After simplification, we get an equation that involves only one variable which can be integrated.

Solving a Differential Equation using Variable Separation

In this section, the speaker explains how to solve a differential equation using variable separation.

Dividing the First Term

To separate the variables, divide the first term by na.

The q variable must be eliminated from this term. Divide it by nothing.

Integrating with Respect to q

Integrate with respect to q.

Use partial fractions to integrate the second term.

Partial Fractions

Use case 1 of partial fractions to find 1 over 2 square root of 2q plus 1 square root of 2q minus 1.

Equate the first denominator to zero and solve for a. Equate the second denominator to zero and solve for b. Substitute these values into the equation and solve for c.

Finding Values of a, b, and c

When q is equal to zero, one equals negative one times negative one times a plus zero times b plus c times negative one multiplied by positive one giving you eight times negative one then for letter b substitute not then one is equal to b times the quantity this value is q which is negative one over square root of two multiplied by square root of two times negative one over square root of two is negative one minus one is negative two so therefore letter b is equal to square root of two over two cross multiply lambda attention because negative times negative is...

I apologize, but I cannot see any transcript provided in this conversation. Please provide me with the transcript so that I can create a comprehensive and informative markdown file as per your requirements.

Solving for s^2

In this section, the speaker explains how to solve for s^2.

Solving for s^2

t times two s squared minus t squared is equal to one times s squared is s squared so since it is

2 t s squared minus e cubed is equal to s squared so by distributing that all then 2 t s squared minus s squared is equal to t cube by transposing that negative t cube and transposing this that s squared then

squared quantity 2t minus 1 equals pq s squared 2t minus 1 equals p cubed so this will be our answer

Example equation

In this section, the speaker provides an example equation.

Example equation

This number 10 page 28 so look at this equation

Separable or variable separable equation

Next method so as you can see this sign x over t is the same as this term number on cosine x over t and then same as with this term number of cosine x over t sine of x over t same function lancilla and this term i made x to

Each term are of the same degree so therefore homogeneous

Substitution

In this section, the speaker explains substitution.

Substitution

X is equal to q t

Is equal to qx

Times two terms so again mass modality multiply not one term multiplied by two terms so your dx there is q dt plus pdq

So this gives you three t cosine x over t multiplied by the value of dx depended on substitution again minus two t sine q t because x is q t same language as in absolute nothing d to say x na

Simplifying

In this section, the speaker explains how to simplify.

Simplifying

So this gives you 3t cosine q times q dt is 3qt cosine qdt then 3t times t is 3t squared multiplied by cosine qdq at the sha minus the quantity 2t sine q dt so that's 2 t sine q dt minus times positive is still minus q 3q t cosine q dt is equal to 0

Simplifying common term

Notice that this first term and last term cancels out three qt cosine qdt and negative three qt cosine q d t is zero so giving you three tsquared cosineqdq minus two tsineqdq

Integration

In this section, the speaker explains integration.

Integration

First term we divide nothing by t squared

Second term an indica line and i sine qsorry by kosha by thank

Sineq dq minus jung p e d by dtsquared eta c sineq divided by sineq is one dt equals zero simplifying further that gives you three integral of cosine q over sine qdq minus integral minus 2 integral of dt over t equals integral of zero let us try to integrate so if you assign q derivative is cosine q dq so that is

Simply ln of sineq times 3 that's 3 l and sineq minus 2 integral of dt over t is ln t and my ln my ln paramas madeleine combine lagindi not in an ln c since ln of c is still a constant so using property and logarithm the 2 becomes the exponent of t that three becomes the exponent of sine so this gives you ln sineq to the third minus l and c transposing this negative on the right side

Gives you sine cubed q is equal to c t squared but then again

General Solution

In this section, the speaker explains how to find a general solution.

General Solution

Q is equal to x over t using our substitution

Finding the Value of B Squared

In this section, the speaker provides a solution to finding the value of b squared.

Solution for Finding B Squared

The answer for finding b squared is "b squared."

This means that if you are asked to find the value of b squared, your answer should be "b squared."