# Thread: Problem with a homogeneous equation

1. Originally Posted by Ackbeet
Once you've recognized the equation as homogeneous, then either v = y/x or v = x/y will render the equation separable. How do I know this? From reading my textbook. There is a proof for this fact in my textbook.
Thats new to me. I didnt realise that either will work for a homogenous equation. Thanks!

2. You're welcome. Incidentally, there is a slight preference as to which substitution to use, merely as a practical matter. Let's say you write your homogeneous equation in the form

M(x,y) dx + N(x,y) dy = 0;

furthermore, let's say M is simpler than N. Then the substitution x = vy will often work slightly better than the other, although you will get a separable equation either way. If N is simple than M, then the substitution y = ux will often work slightly better. See Zill's A First Course in Differential Equations, 6th Ed., p. 55.

3. Originally Posted by Ackbeet
You're welcome. Incidentally, there is a slight preference as to which substitution to use, merely as a practical matter. Let's say you write your homogeneous equation in the form

M(x,y) dx + N(x,y) dy = 0;

furthermore, let's say M is simpler than N. Then the substitution x = vy will often work slightly better than the other, although you will get a separable equation either way. If N is simple than M, then the substitution y = ux will often work slightly better. See Zill's A First Course in Differential Equations, 6th Ed., p. 55.

If we say that $M(x,y) dx + N(x,y) dy = 0$ is homogeneous from order $n$ then we can write:

$M(x,y) dx + N(x,y) dy = x^n \{ M_1 (\frac{y}{x})dx+N_1(\frac{y}{x})dy \}=0$

Or:

$M_1 (\frac{y}{x})dx+N_1(\frac{y}{x})dy =0$

Now, the substitution $y=vx$, $dy=vdx+xdv$ will give:

$M_1(v)dx+N_1(v)(vdx+xdv)=0$

Or:

$(M_1(v)+vN_1(v))dx+xN_1(v)dv=0$

Or separable equation:

$\frac{dx}{x}+\frac{N_1(v)dv}{M_1(v)+vN_1(v)}=0$

Page 2 of 2 First 12