What Steps Are Misunderstood in Solving Non-Linear Homogeneous Equations?

[fireandwater]

Hi. I'm having a problem understanding how to solve non-linear homogeneous equations. For example, (x$$^{}2$$+y$$^{}2$$)dx +xydy = 0 ; x=1, y=1

I understand that y=xv, v=y/x, and dy=xdv +vdx

To sub in,
x$$^{}2$$ + (v$$^{}2$$x$$^{}2$$)dx +x$$^{}2$$v(vdx+xdv) = 0

Here's where I get lost:
x$$^{}2$$[(1+v$$^{}2$$)dx +v$$^{}2$$dx +xvdv] = 0 =>

x$$^{}2$$(1+v$$^{}2$$)dx +v$$^{}2$$dx+xvdv = 0 =>

1+2v$$^{}2$$dx +xvdv = 0 =>

1+2v$$^{}2$$dx = -xvdv =>

$$\int-dx/x$$ = int(vdv/1+2v^2) =>

-ln x = 1/4 ln (1+2v$$^{}2$$) +C =>

ln x$$^{}4$$ + ln(1+2v$$^{}2$$) +4c = A

I think I'm just getting lost in all the algebraic "cleaning up", but I can't figure it out. Can someone pick it apart for me?

 

[arildno]

Hi. I'm having a problem understanding how to solve non-linear homogeneous equations. For example, (x$$^{}2$$+y$$^{}2$$)dx +xydy = 0 ; x=1, y=1

I understand that y=xv, v=y/x, and dy=xdv +vdx

To sub in,
x$$^{}2$$ + (v$$^{}2$$x$$^{}2$$)dx +x$$^{}2$$v(vdx+xdv) = 0

THis should read:
$$x^{2}dx+v^{2}x^{2}dx+x^{2}v(vdx+xdv)=0$$

Clean this up as follows:
$$(x^{2}+v^{2}x^{2}+x^{2}v^{2})dx+x^{3}vdv=0$$
That is to say:
$$x^{2}(1+2v^{2})dx+x^{3}vdv=0$$
or, divided by x^{2}:
$$(1+2v^{2})dx+xvdv=0$$

See if you follow this so far!

 

[tacman]

It looks like you have the right approach, but there are a few small errors in your steps. Let's go through it together to see where you may have gone wrong.

First, we start with the given equation: (x^{}2+y^{}2)dx +xydy = 0

Next, we use the substitution y=xv, v=y/x, and dy=xdv +vdx to get:

x^{}2 + (v^{}2x^{}2)dx +x^{}2v(vdx+xdv) = 0

Now, we can simplify the equation by factoring out an x^{}2:

x^{}2[(1+v^{}2)dx +v^{}2dx +xvdv] = 0

Note that you have a small error here, as you have written x^{}2v(vdx) instead of x^{}2v(xdv). This is just a small mistake, but it's important to be careful with these substitutions.

Next, we can combine like terms inside the brackets to get:

x^{}2(1+v^{}2)dx +x^{}2v(vdx) = 0

Again, there is a small error here, as you have written x^{}2v(vdx) instead of x^{}2v(xdv).

Now, we can use the substitution v=y/x to get:

x^{}2(1+v^{}2)dx +xyv(xdv) = 0

Note that we can simplify this further by dividing both sides by x:

x(1+v^{}2)dx +yv(xdv) = 0

Now, we can integrate both sides with respect to x to get:

∫x(1+v^{}2)dx + ∫yv(xdv) = ∫0dx

The right side is just a constant, so we can ignore it. For the left side, we can use the power rule to integrate x(1+v^{}2)dx, which gives us:

(1/2)x^{}2 + (1/2)x^{}2v^{}2 + C = 0

Now, we can use the substitution v=y/x to get:

(1/2)x^{}2 + (1/2)y^{}2 + C