Question

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Answer 1

SEE THE ATTACHMENT FOR DETAILED SOLUTION TO THE QUESTION DUDE...❤❤

ANOTHER PROCESS::--

It's an exact equation, the left member is the differential of the function:
F(x,y)=x2+3xy−5x−y2/2−2yF(x,y)=x2+3xy−5x−y2/2−2y.
So the equation is dF=0dF=0 which solution is F=CF=C where C is a constant:
x2+3xy−5x−y2/2−2y=Cx2+3xy−5x−y2/2−2y=C.
USING ANOTHER WAY (“substitution”?)
(2x+3y−5)dx+(3x−y−2)dy=0(2x+3y−5)dx+(3x−y−2)dy=0
x=u+ax=u+a
y=v+by=v+b
2x+3y−5=2u+2a+3v+3b−52x+3y−5=2u+2a+3v+3b−5
I impose that the terms not having the variables uu and vv to be zero:
2a+3b−5=0(1)(1)2a+3b−5=0
3x−y−2=3u+3a−v−b−23x−y−2=3u+3a−v−b−2
as before:
3a−b−2=0→b=3a−23a−b−2=0→b=3a−2
substitute in (1):
2a+9a−6−5=0→a=1−−>b=1

Answer 2

Let 3x-y-2=u thus dy=3dx-du. Then equation becomes:

(2x+9x-3u-6–5)dx+u(3dx-du)=0 or 11(x-1)dx=udu Let x-1=v, thus 11vdv=udu

If we integrate, (11/2)*V^2=(1/)2*u^2+k(0)/2 or u=+-((11x-1)^2+k(1))^0.5 or

y=3x-2+-(11x^2–22x+k)^0.5