Question

5. (x + 2y − 1)dx − (2x + y − 5)dy = 0

Answer 1

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 -

$\dfrac{x^2}{2}+2xy-x+\dfrac{y^2}{2}-5y=c$