(2x+5y+2)dx+(10y-4x-4)dy=0.
This can also be written: (2(x+1)+5y)dx+(10y-4(x+1))dy=0.
Let X=x+1 and Y=y. We now have a homogeneous DE: (2X+5Y)dX-(4X-10Y)dY=0;
let Y=vX where v is a function of X, so dY/dX=v+Xdv/dX
dY/dX=(2X+5Y)/(4X-10Y))=(2X+5vX)/(4X-10vX))=(2+5v)/(4-10v)=v+Xdv/dX
Xdv/dX=(2+5v)/(4-10v)-v=(2+v+10v2)/(4-10v)
∫(4-10v)dv/(10v2+v+2)=∫dX/x.
-½∫(20v+1-9)dv/(10v2+v+2)=ln(ax), where a is a constant.
-½ln(10v2+v+2)+(9/2)∫dv/(10v2+v+2)=ln(ax).
10v2+v+2=10(v2+0.1v+0.2)=10(v2+0.1v+0.0025-0.0025+⅔)=10((v+0.05)2+797/1200). Let b2=797/1200=0.6642 approx.
Let v+0.05=btanθ, dv=bsec2θdθ;
10((v+0.05)2+b2)=10(b2tan2θ+b2)=10b2sec2θ.
(9/2)∫dv/(10v2+v+2)=(9/2)∫bsec2θdθ/10b2sec2θ=(0.45/b)∫dθ=(0.45/b)θ=
(0.45/b)tan-1((v+0.05)/b).
-½ln(10v2+v+2)+(0.45/b)tan-1((v+0.05)/b)=ln(ax).
-½ln(10(Y/X)2+(Y/X)+2)+(0.45/b)tan-1((Y/X+0.05)/b)=ln(ax),
-½ln(10y/(x+1)2+y/(x+1)+2)+(0.45/b)tan-1((y/(x+1)+0.05)/b)=ln(ax).
In this equation b has the value 0.6642 approx. The equation can be expressed in different forms, including mixed exponential and trigonometric expressions.