(2x+3y-1)dx + (2x+3y-5)dy = 0 variable separable

find the DE.
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24 Answers

dy/dx=-2x+3y-1/2x+3y-5
by Level 12 User (101k points)
(1) now let 2x+3y-1
by Level 12 User (101k points)
→ ø(x)
by Level 12 User (101k points)
Ø (x) _= Ø
by Level 12 User (101k points)
So that 2+3y'(x) → Ø'(x)
by Level 12 User (101k points)
Therefore dy/dx→ 1/3( dø/ dx -2)
by Level 12 User (101k points)
Then (1) is transformed into :
by Level 12 User (101k points)
dØ/dx-2= -3Ø/Ø-4
by Level 12 User (101k points)
dØ/dx=- 3Ø/Ø-4
by Level 12 User (101k points)
dø/dx=8+ø/4-ø
by Level 12 User (101k points)
Therefore:
by Level 12 User (101k points)
{4-ø/8+ø dø={ dx
by Level 12 User (101k points)
→ 4{ d(8+ø)/8+ø -{ødø/8+ø
by Level 12 User (101k points)
=x+c ,
by Level 12 User (101k points)
4 ln(8+ø)-[(8+ø)-8 ln(8+ø)]
by Level 12 User (101k points)
13 ln(8+ø)
by Level 12 User (101k points)
X+c+(8+ø)
by Level 12 User (101k points)
Ø→2x+3y-1
by Level 12 User (101k points)
So we get 12 ln(3y+2x+7)
by Level 12 User (101k points)
=3(x+y)+(7+c)
by Level 12 User (101k points)
ln(3y+2x+7)
by Level 12 User (101k points)
=1/4(x+y)+ c ~ upper
by Level 12 User (101k points)
3y+2x+7=c base 0•e^1/4(x+y
by Level 12 User (101k points)
(Implicit form
by Level 12 User (101k points)

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