# Solve Homogeneous D.E. integrating

• Jtechguy21
In summary, the conversation discusses how to solve the differential equation dy/dx = (y-x)(y+x) by substituting y=ux and dy=udx+xdu. After simplifying, the equation is put in separable form and the integration is completed to find the solution.
Jtechguy21

## Homework Statement

Dy/Dx = (Y-x)/(Y+x)

Y=ux
dy=udx+xdu

## The Attempt at a Solution

Dy/Dx = (Y-x)/(Y+x)

Plug in my substitutions
udx+xdu(1/dx)=(ux/ux+x) - X/(ux+x)

Simplify
u+x(du/dx)=(ux)/x(u+1) - (x)/((x)(u+1))

u+x(du/dx)=u/(u+1) -(1)/(u+1))

u+x(du/dx)=u-1/(u+1)

This is where I think i begin to mess up

u+du=(u-1)/(u+1) dx/x

substract (u-1)/(u+1) to the other side

u-(u-1)/(u+1) du=dx/x

I know the right side integrates to Lnx +c

but on the left side if i do
(u^2-1)/(u+1)

I split it up into

the integral (u^2)/(u+1) minus integral of 1/(u+1)
(u^2)/(u+1)<-use long division

I get u+(1/u+1) minus the integral of 1/(u+1)
i am left with just the integral
of u

u^2/2= lnx+c

plug u back in.

((y/x)^2)/2 =lnx +c

is this sufficient of an answer?
according to the answer key I am going to end up with the arctan somewhere in my answer. so i may have already messed up :(

Last edited:
Jtechguy21 said:

## Homework Statement

Dy/Dx = (Y-x)(Y+x)

Y=ux
dy=udx+xdu

## The Attempt at a Solution

Dy/Dx = (Y-x)(Y+x)

Plug in my substitutions
udx+xdu(1/dx)=(ux/ux+x) - X/(ux+x)

Why are you dividing by $y + x$ here? Is your equation actually
$$\frac{dy}{dx} = \frac{y - x}{y + x}$$
and not
$$\frac{dy}{dx} = (y - x)(y + x)$$
as you have written?

Simplify
u+x(du/dx)=(ux)/x(u+1) - (x)/((x)(u+1))

u+x(du/dx)=u/(u+1) -(1)/(u+1))

u+x(du/dx)=u-1/(u+1)

This is where I think i begin to mess up

u+du=(u-1)/(u+1) dx/x

This should be "u dx/x + du" on the left hand side.

substract (u-1)/(u+1) to the other side

You can't; it's multiplied by dx/x.

What you have after replacing $y$ is
$$x\frac{du}{dx} + u = \frac{u - 1}{u + 1}$$
Subtracting $u$ from both sides and then dividing by $x$ puts this in the separable form
$$\frac{du}{dx} = \frac1x \left(\frac{u-1}{u+1} - u\right)$$
Continue.

## 1. What is a homogeneous differential equation?

A homogeneous differential equation is a type of differential equation where all the terms contain the dependent variable and its derivatives. This means that the equation can be written in a form where the dependent variable and its derivatives are the only variables present.

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

To solve a homogeneous differential equation, you can use the technique of separation of variables or the method of substitution. First, you need to rearrange the equation so that all the terms involving the dependent variable and its derivatives are on one side, and all the other terms are on the other side. Then, you can integrate both sides and solve for the dependent variable.

## 3. What is the importance of integrating a homogeneous differential equation?

Integrating a homogeneous differential equation allows us to find the general solution, which is a function that satisfies the equation for all possible values of the independent variable. This is important because it helps us understand the behavior of the system described by the differential equation and make predictions about its future behavior.

## 4. Can a homogeneous differential equation have more than one solution?

Yes, a homogeneous differential equation can have infinitely many solutions. This is because the general solution contains a constant of integration, which can take on any value. However, for a specific initial condition, there will be a unique solution.

## 5. What are some real-life applications of solving homogeneous differential equations?

Homogeneous differential equations are commonly used in physics, engineering, and economics to model various systems, such as population growth, radioactive decay, and heat transfer. They are also used in the field of control theory to analyze and design systems that can be described by differential equations.

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