# First order ODE, The Homogeneous Method.

• knowlewj01
In summary, the conversation revolved around solving an equation involving fractions and substitutions. The final solution involved rewriting the equation as a rational expression and using algebraic techniques to separate the variables, allowing for integration to find the solution.
knowlewj01

## Homework Statement

$$\frac{1}{xy}$$ $$\frac{dy}{dx}$$ = $$\frac{1}{(x^2 + 3y^2)}$$

## Homework Equations

used the substitutions:

v = $$\frac{x}{y}$$ ,and
$$\frac{dy}{dx}$$ = $$v + x$$ $$\frac{dv}{dx}$$

## The Attempt at a Solution

took out a factor of xy on the denominator of the term on the right hand side and multiplied through by xy, made the substitutions and at the moment I have something that looks like this:

$$\frac{1}{v + 3/v}$$ = v + x $$\frac{dv}{dx}$$

as it stands I can't see a way to do this by separation of variables, is it solvable by integrating factor? anyone got any ideas?

You used the substitution v = x/y, but your work in calculating dy/dx looks like you started with y = vx, which is not equivalent to v = x/y.

ah thanks, i'll give it another go.

I was able to get the equation separated with the substitution v = y/x.

I have the substitutions:

v = $$\frac{y}{x}$$

$$\frac{dy}{dx}$$ = v + x $$\frac{dv}{dx}$$

so now my equation looks a bit like this:

$$\frac{1}{1/v + 3v}$$ = v + x $$\frac{dv}{dx}$$

I played with it a bit and still can't see how to separate it, can anyone point me in the right direction?

thanks

What you have is fine, but needs cleaning up using ordinary algebra techniques.

1. Rewrite 1/(1/v + 3v)) as a rational expression - one numerator and one denominator, with no fractions in the top or bottom.
2. Subtract v from both sides, so that you have x*dv/dx on one side, and a rational expression on the other.
3. Divide both sides by x. You should now be able to move dv or dx so that everything in v or dv is on one side, and everything in x or dx is on the other. The equation is now separated, and you can integrate to find the solution.

## 1. What is a first order ODE?

A first order ODE (ordinary differential equation) is a mathematical equation that involves a function and its first derivative. It describes how a variable changes over time or space in relation to its own value and the values of other variables.

## 2. What is the Homogeneous Method for solving first order ODEs?

The Homogeneous Method is a technique used to solve first order ODEs that are homogeneous, meaning that all terms in the equation involve the dependent variable and its derivatives. It involves substituting a new variable for the dependent variable and rearranging the equation to make it separable, then integrating to find the solution.

## 3. How is the Homogeneous Method different from other methods for solving first order ODEs?

The Homogeneous Method is specific to solving homogeneous first order ODEs. Other methods, such as the separation of variables method or the integrating factor method, can be used to solve non-homogeneous first order ODEs. The Homogeneous Method is often preferred because it is more straightforward and does not require the use of integrating factors.

## 4. When is the Homogeneous Method most useful?

The Homogeneous Method is most useful when solving first order ODEs that are homogeneous and have initial conditions provided. It is also beneficial when the equation is separable, meaning it can be divided into two parts that only involve one variable each. In these cases, the Homogeneous Method can provide a quick and efficient solution.

## 5. Are there any limitations to using the Homogeneous Method?

While the Homogeneous Method is a useful technique for solving first order ODEs, it does have its limitations. It can only be used for homogeneous equations, so it cannot be applied to non-homogeneous equations. Additionally, it may not be suitable for equations with complex or non-separable terms, in which case other methods may be more effective.

• Calculus and Beyond Homework Help
Replies
19
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
384
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
25
Views
684
• Calculus and Beyond Homework Help
Replies
11
Views
927
• Calculus and Beyond Homework Help
Replies
2
Views
690
• Calculus and Beyond Homework Help
Replies
21
Views
1K
• Calculus and Beyond Homework Help
Replies
33
Views
3K
• Calculus and Beyond Homework Help
Replies
2
Views
580