Mamats po sana po matulungan po ninyo ako kasi dyan po ako nastuck. Sana nandito rin si sir electrons. Salamat po!
sir check mo ulit if meron tama na ba... konti nalang kasi natatandaan ko about D.E ehh.. pero check mo if wala na bang error...
(2x+y+6)dx + (2x+y)dy = 0 ; y = vx ; dy = vdx + xdv
(2x+y)(dx+dy) = -6dx
-6dx / (2x+y) = dx+dy then -6 / (2x+y) = (dx+dy) / dx then
subs. y=vx and dy = vdx + xdv... the eq. will become:
-6 / x(2+v) = (1+v) + xdv/dx multiply both sides by x..
-6x / (2+v) = x(1+v) + (x²dv/dx) transpose x(1+v) on the left eq...
shortcut na lang sir...
-x[(6 / (2+v)) + (1=v)] = x²dv / dx then cross multiply....
the eq will be..
-xdx / x² = (2+v)dv / (v²+3v+8) then the integral of (-x / x²)dx is:
-ln|x| and the integral of the other eq. is..
[(2v+3) / 2(v²+3v+8)] + [1 / 2(v²+3v+8)](2v+3) / 2(v²+3v+8) will become ½ln(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..
½ln(v²+3v+8) + [tan^-1 [(2v+3) / √23)] / √23] + c = 0
½ln[(x²/y²) + (3x/y) +8] + tan^-1[((2x/y)+3)/√23)] / √23] + c = 0
combine both terms..
ln|x| + ½ln[(x/y)² + (3x/y) +8] + tan^-1[((2x/y)+3)/√23)] / √23] + c = 0
yan po yung revised solution ko, check nyu na lang ulit, if may mali, si sir electrons na lang gagawa,, ahahaha... thank you...