5. (x + 2y − 1)dx − (2x + y − 5)dy = 0
The give differential equation is-
(x + 2y − 1)dx − (2x + y − 5)dy = 0
first, re-arrange the problem to be-
"x+2y-1+(2x+y-5)\\dfrac{dy}{dx}=0"
let "M=x+2y-1" and "N=2x+y-5" and define "Fx=M" and "Fy=N"
Differentiate M w.r.t y-
"M_y=2"
Differentiate N w.r.t. x-
"N_x=2"
As "M_y=N_x"
thus the problem is an EXACT ODE
fourth, find integral of M dx holding y constant "= \\dfrac{x^2}{2}+2xy-x+h(y) = F"
find "F_y= 2x+h'= N = 2x+y-5 ..." ' solve for "h'=y-5" and "h=\\dfrac{y^2}{2}-5y"
Complete solution is -
"\\dfrac{x^2}{2}+2xy-x+\\dfrac{y^2}{2}-5y=c"
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