persamaan differensial

Question

(2x+5y+2)dx+(10y-4x-4)dy=0

Answer 1

(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.