Homogeneous differential equation

B

burt

Full Member
Joined
Aug 1, 2019
Messages
254
Why do we use [MATH]\frac{y}{x}[/MATH] to get the general solution of a homogeneous differential equation? Why not [MATH]\frac{x}{y}[/MATH]?
 
burt said:
Why do we use [MATH]\frac{y}{x}[/MATH] to get the general solution of a homogeneous differential equation? Why not [MATH]\frac{x}{y}[/MATH]?
Click to expand...

Suppose we have an ODE of the form:

[MATH]\d{y}{x}=f\left(\frac{y}{x}\right)[/MATH]
When we use the substitution:

[MATH]u=\frac{y}{x}\implies y=ux\implies \d{y}{x}=u+x\d{u}{x}[/MATH]
We may now write the ODE as:

[MATH]u+x\d{u}{x}=f(u)[/MATH]
We now have a separable equation:

[MATH]\int\frac{1}{f(u)-u}\,du=\int \frac{1}{x}\,dx[/MATH]
Now suppose we choose to write:

[MATH]\d{y}{x}=f\left(\frac{x}{y}\right)[/MATH]
[MATH]u=\frac{x}{y}\implies y=\frac{x}{u}\implies \d{y}{x}=\frac{u-x\d{u}{x}}{u^2}[/MATH]
And our ODE becomes:

[MATH]\frac{u-x\d{u}{x}}{u^2}=f(u)[/MATH]
[MATH]x\d{u}{x}=u-u^2f(u)[/MATH]
[MATH]\int \frac{1}{u-u^2f(u)}\,du=\int \frac{1}{x}\,dx[/MATH]
As we can see, both substitutions will lead to a separable ODE, but the first will likely be simpler. However, if you think that for a particular ODE the second substitution will more easily lead to a solution, then there's no reason you cannot use it.
 
You must log in or register to reply here.
Top