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

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SUMMARY

This discussion focuses on solving non-linear homogeneous equations, specifically the equation (x² + y²)dx + xydy = 0 with initial conditions x=1 and y=1. The transformation y = xv and v = y/x is utilized, leading to the substitution that simplifies the equation to (1 + 2v²)dx + xvdv = 0. The participants emphasize the importance of correctly manipulating algebraic expressions and integrating to arrive at the solution, which involves logarithmic functions.

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

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

To sub in,
x[tex]^{}2[/tex] + (v[tex]^{}2[/tex]x[tex]^{}2[/tex])dx +x[tex]^{}2[/tex]v(vdx+xdv) = 0

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

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

1+2v[tex]^{}2[/tex]dx +xvdv = 0 =>

1+2v[tex]^{}2[/tex]dx = -xvdv =>

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

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

ln x[tex]^{}4[/tex] + ln(1+2v[tex]^{}2[/tex]) +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?
 
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fireandwater said:
Hi. I'm having a problem understanding how to solve non-linear homogeneous equations. For example, (x[tex]^{}2[/tex]+y[tex]^{}2[/tex])dx +xydy = 0 ; x=1, y=1

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

To sub in,
x[tex]^{}2[/tex] + (v[tex]^{}2[/tex]x[tex]^{}2[/tex])dx +x[tex]^{}2[/tex]v(vdx+xdv) = 0
THis should read:
[tex]x^{2}dx+v^{2}x^{2}dx+x^{2}v(vdx+xdv)=0[/tex]

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

See if you follow this so far!
 

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