Solving a homogenous equation

Solve the homeogenous euqation (x+y)dx + (y-x) dy =0

(y-x)dy= -(x+y)dx

dy/dx= -(x+y)/(y-x)

dy/dx -x/(y-x) - y/(y-x)...

Now I think we're supposed to make u= y/x or something like that- whatever fits....

butI don't know what to do

mathstresser said:

Now I think we're supposed to make u= y/x or something like that...

Click to expand...

You already have a perfectly good approach there. Just need to fill in the details.

\(\displaystyle \L u= \frac{y}{x}\) gives \(\displaystyle y=xu\) and

\(\displaystyle \L \frac{dy}{dx}=u + x \frac{du}{dx}\)

Now write your DE in terms of u,

\(\displaystyle \L \frac{dy}{dx}= -\frac{y+x}{y-x}\)

\(\displaystyle \L u + x \frac{du}{dx} = -\frac{xu+x}{xu-x} = -\frac{u+1}{u-1}\)

\(\displaystyle \L x \frac{du}{dx} = -\frac{u+1}{u-1} - u = \frac{1+u^2}{1-u}\)

Now the equation is separable,

\(\displaystyle \L \frac{1-u}{1+u^2} du = \frac{1}{x} dx\)

You can finish it up from here.

Hello, mathstresser!

I'll show you the way I was taught to do these.
It may make things clearer . . . or not.

Solve the homeogenous equation: \(\displaystyle (x\,+\,y)dx\,+\,(y\,-\,x) dy \:=\:0\)

Click to expand...

Rearrange and we get: \(\displaystyle \L\:\frac{dy}{dx}\:=\:\frac{x\,+\,y}{x\,-\,y}\)

Divide top and bottom by \(\displaystyle x:\) \(\displaystyle \L\:\frac{dy}{dx}\:=\:\frac{1\,+\,\frac{y}{x}}{1\,-\,\frac{y}{x}}\)


Let \(\displaystyle u\,=\,\frac{y}{x}\;\;\Rightarrow\;\;y\:=\:ux\;\;\Rightarrow\;\;\frac{dy}{dx}\:=\:u\.+\,x\frac{du}{dx}\)

Substitute: \(\displaystyle \L\:u\,+\,x\frac{du}{dx}\;=\;\frac{1\,+\,u}{1\,-\,u}\)

Then: \(\displaystyle \L\:\frac{du}{dx}\;=\;\frac{1\,+\,u}{1\,-\,u}\,-\,u\;\;\Rightarrow\;\;\frac{du}{dx}\:=\:\frac{1\,+\,u^2}{1\,-\,u}\)

Separate variables: \(\displaystyle \L\:\frac{1\,-\,u}{1\,+\,u^2}\,du\;=\;\frac{dx}{x}\)

. . the same result as mark07 . . .

Thank you!