# Homogeneous Differential Equation

• BesselEquation
In summary, the conversation is about solving a specific differential equation and the solver is having trouble solving an integral that arises in the process. They discuss using a standard method and Wolfram's solution, but disagree on the final step of the integration.
BesselEquation

## Homework Statement

Solve the following differential equation:
y' = y / [ x + √(y^2 - xy)]

2. The attempt at a solution
Using the standard method for solving homogeneous equations, setting u = y/x, I arrive at the following:

± dx/x = [1±√(u^2-u) ]/ [u√(u^2-u)] which in turn, I get the following integral after simplifying:

∫du/[u√(u^2-u)], seems quite unsolvable to me...

Note that$$\frac{1}{u\sqrt{u^2-u}}=\frac{1}{u^{3/2}(u-1)^{1/2}}$$Here's where the guessing and experience with derivatives comes in. The idea is to write this as a perfect differential which would make the integration trivial. You know that the term ##u^{3/2}## in the denominator arises by taking the derivative of ##1/u^{1/2}## and the term ##(u-1)^{1/2}## in the denominator arises by taking the derivative of ##(u-1)^{1/2}##. You also know the product rule of differentiation. What do you get when you take the derivative of ##\frac{(u-1)^{1/2}}{u^{1/2}}~##?

Delta2 said:
Well , according to wolfram the integral $$\int \frac{du}{u\sqrt{u^2-u}}=2\sqrt{\frac{u-1}{u}}$$
$$\int \frac{du}{u\sqrt{u^2-u}}=\int \frac{du}{u^2\sqrt{1-1/u}}=\int \frac{d(1-1/u)}{\sqrt{1-1/u}}=\int 2d\sqrt{1-1/u}$$

vela, MathematicalPhysicist and Delta2
Woah alright, great...

lurflurf said:
$$\int \frac{du}{u\sqrt{u^2-u}}=\int \frac{du}{u^2\sqrt{1-1/u}}=\int \frac{d(1-1/u)}{\sqrt{1-1/u}}=\int 2d\sqrt{1-1/u}$$

This is not quite right; it should be
$$\int \frac{d\:\text{anything}}{\sqrt{\text{anything}} }= 2 \sqrt{ \text{anything}}$$

Ray Vickson said:
This is not quite right; it should be
$$\int \frac{d\:\text{anything}}{\sqrt{\text{anything}} }= 2 \sqrt{ \text{anything}}$$
I don't understand where you disagree, isn't it ##\frac{d f}{\sqrt{f}}=2d(\sqrt{f)}## and then the integral operator cancels out with the differential operator d and leaves just ##2\sqrt{f}##

Delta2 said:
I don't understand where you disagree, isn't it ##\frac{d f}{\sqrt{f}}=2d(\sqrt{f)}## and then the integral operator cancels out with the differential operator d and leaves just ##2\sqrt{f}##
OK, you are right. Basically, I did not notice that you just left out the last step.

## 1. What is a homogeneous differential equation?

A homogeneous differential equation is a type of differential equation where all terms can be expressed as a function of the dependent variable and its derivatives. In other words, the equation is "homogeneous" in the sense that all terms have the same degree.

## 2. How do you solve a homogeneous differential equation?

To solve a homogeneous differential equation, you can use the method of separation of variables, where you separate the variables and integrate both sides of the equation. Another method is to use substitution, where you substitute a new variable to transform the equation into a separable form.

## 3. What is the difference between a homogeneous and non-homogeneous differential equation?

The main difference between a homogeneous and non-homogeneous differential equation is that the former only contains terms with the same degree, while the latter may have terms with different degrees. This makes solving homogeneous differential equations easier compared to non-homogeneous ones.

## 4. What are some real-world applications of homogeneous differential equations?

Homogeneous differential equations have various applications in physics, engineering, and economics. For example, they can be used to model population growth, chemical reactions, and electric circuits. They are also commonly used in the study of fluid dynamics and heat transfer.

## 5. Can a non-homogeneous differential equation be transformed into a homogeneous one?

Yes, a non-homogeneous differential equation can be transformed into a homogeneous one by using a suitable substitution. This is known as the method of reduction of order. However, the resulting homogeneous equation may have additional solutions that do not satisfy the original non-homogeneous equation.

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