₦airaland Forum

Richiez: i think there is an error in one of the signs, but since it's for self understanding, let me solve this
(2x-4y+5)dy/dx -x+2y+3 =0
(2x-4y+5)dy/dx =x-2y-3
dy/dx = (x-2y-3)/2x-4y+5 ..........(1)
let us use the substitution y = v + x/2. this substitution is chosen so as to replace all y and x terms in the R.H.S with v term
now let's watch out!
if y= v+x/2
dy/dx = dv/dx +1/2
eqn(1) becomes

dv/dx+1/2=
x-2(v+x/2)-3/[2x-4(v+x/2)+5]

dv/dx +1/2 = x-2v-x-3/(2x-4v-2x+5)

dv/dx = -2v-3/(5-4v) -1/2

dv/dx = -4v-6-5+4v/[2(5-4v)]

dv/dx = -11/(10-8v)

by separating the variables,
(10-8v)dv = -11dx

integrating both sides
10v - 4v² = -11x + c

but recall that y= v+x/2
therefore v= y-x/2
hence the real soln to the diff eqn is
10(y - x/2) - 4(y - x/2)² = -11x +c
from here, you can expand the terms in the bracket and re-arrange

personal59: pls help me to solve this ordinary differential equation (2x-4y+5)dy/dx+x-2y+3 equal 0 pls d calculation (steps) is needed for self understanding thanks.

Richiez: solution loading...
check back by 2pm

Richiez: This thread is the meeting point for nairaland maths gurus...i dare you all to ask me a question in mathematics without me having an answer to them, LETS START

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