Patulong Guys about D.E.
homogenous function po kmi.
heto po problem:
(2x+y+6)dx + (2x+y)dy = 0
substitute y to vx
salamat po!
sir check mo if tama yung solution ko: shortcut na lang yung sol'n.
y = vx ; dy = vdx + xdv
(2x+vx)dx + 6dx + (2x²+vx²)dv + (2vx + v²x)dx = 0
x(2+v)dx + 6dx + x(2x+vx)dv + x(2v+v²)dx = 0 then divide eq. by x
(2+v)dx +6dx +(2x+vx)dv + (2v+v²)dx = 0 then combine similar terms
(v²+3v+8)dx + x(2+v)dv = 0 then multiply the whole equation by:
1 / x(v²+3v+8) resulting eq. will be:
dx/x + (2+v)dv / (v²+3v+8) = 0integrate the equation:
integral of dx/x = log(x) ; integral of the (2+v)dv / (v²+3v+8) is:
rewrite the integrand as:
[(2v+3) / 2(v²+3v+8)] + [1 / 2(v²+3v+8)] dv short cut na lang to sir.. expand mo na lang.. ehe then the next step is to integrate term by term..
(2v+3) / 2(v²+3v+8) will become ½log(v²+3v+8) then integrate the next term..
complete the square of 1 / (2(v²+3v+8) it will become like this:
(v+ 3/2)² + (23/4) let u = (v+(3/2)) ; du = dv
½∫1 / (u² + (23/4) then the integral is equal to:
[2tan^-1 (2u/ √23)] / √23 then subs. u = (v + 3/2)
[2tan^-1 [(2(v+ 2/3) / √23)] / √23] then simplify it.. pag na simplify mo na
combine yung mga na integrate mo na term..
log(x) + ½log(v²+3v+8) + [tan^-1 [(2v+3) / √23)] / √23] + c = 0
then subs. v = x/y
eto yugn final answer ko, pero d ako sure kung ito nga ba talaga yung sagot...
log(x) + ½log[(x²/y²) + (3x/y) +8] + tan^-1[((2x/y)+3)/√23)] / √23] + c = 0
check mo na lang po kung tama..