Question

solve y(2x+y-2)dx=2(x+y)dy

Answer 1

Solution:

On comparing given equation with $M(x,y)dx + N(x,y)dy =0$ , we get,

$M(x,y) = y(y + 2x -2 ) , N(x,y) = -2(x + y)$

The equation is not exact because

$M_y =2(x + y -1) ; N_x = -2.$ But $( M_y - N_x )/N =- 1.$

So the $I.F. = e^{-x}$

may be used to obtain the exact equation $P(x,y)dx + Q(x,y)dy = 0$

with

$P(x,y) = (e^{-x})(y(y+2x - 2)) , Q(x,y) = - 2e^{-x}(x+y)$ and $P_y = Q_x = e^{-x} 2( y+x-1).$ The equation is an exact differential dF(x,y) = 0 then, the solution

is F(x,y) = C, with $F_x =P = (e^{-x})( y( y +2x - 2))$ ...(1) ,

$F_y = Q = - 2e^{-x}(x + y)$ ...(2).

Integrating eq. (1) yields $F(x,y) = -e^{-x} y( y + 2x) + G(x)$ and using eq.(2) gives

G’(x) =0 so G(x) =C .

Then the solution is : $F(x,y) = - e^{-x} y (y+2x) = C$