Derive the Integrating Factor for Homogeneous DE

[lkh1986]

Homework Statement

I have this statement:
If $$M(x,y)dx+N(x,y)dy=0$$ is a homogeneous DE, then $$μ(x,y)=\frac{1}{xM+yN}$$ is its integrating factor. The problem is, how do we derive this integrating factor?

Homework Equations

For homogeneous DE, we have $$f(kx,ky)=k^n*f(x,y)$$
We also have $$\frac{dy}{dx}=-\frac{M(x,y)}{N(x,y)}=-\frac{M(x,xv)}{N(x,xv)}=-\frac{x^pM(1,v)}{x^pN(1,v)}=-\frac{M(1,v)}{N(1,v)}=F(v)=F(\frac{y}{x})$$

$$\frac{M(x,y)}{N(x,y)}dx+dy=0$$ becomes
$$-F(\frac{y}{x})dx+dy=0$$

The Attempt at a Solution

I try to introduce $$μ$$ into the original DE, then I try to derive the factor, which I know the final answer would be $$μ(x,y)=\frac{1}{xM+yN}$$, but I get very complicated formula, which I cannot simplify. I suspect that there're more properties for homogeneous DE?

Thanks.

 

[mighty2000]

Hi there,

To derive the integrating factor for a homogeneous DE, we can use the following steps:

1. Rewrite the DE in the form of \frac{dy}{dx}=-\frac{M(x,y)}{N(x,y)}.

2. Substitute \frac{y}{x} with a new variable, let's say v. This will give us \frac{dy}{dx}=F(v).

3. To make the equation homogeneous, we can divide both sides by x^n, where n is the highest power of x in the equation. This will give us \frac{1}{x^n}\frac{dy}{dx}=F(v).

4. Now, we can rewrite the equation as \frac{1}{x^n}dy=F(v)dx.

5. We can then introduce the integrating factor μ, which is a function of x and y, into the equation. This will give us \mu\frac{1}{x^n}dy=\mu F(v)dx.

6. We can then rearrange the equation to get \frac{d}{dx}(\mu y)=\mu F(v).

7. Integrating both sides with respect to x, we get \mu y=\int \mu F(v) dx + C, where C is a constant of integration.

8. Simplifying the integral on the right hand side, we get \mu y=\int \frac{\mu}{x^n}F(v)dx + C.

9. Comparing this to the original equation, we can see that the integrating factor is μ(x,y)=\frac{1}{x^n F(v)}.

10. Finally, substituting back in our variable v, we get μ(x,y)=\frac{1}{x^n F(\frac{y}{x})}=\frac{1}{xM+yN}.

I hope this helps! Let me know if you have any further questions.