Question

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Answer 1

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(2x+3y-5)dy/dx=-(3x+2y-5)

⇒ (2x+3y-5)dy=-(3x+2y-5)dx

then by compering with

mdy+ndx=0

⇒ m=(2x+3y-5) and n=(3x+2y-5)

dm/dx=2 and dn/dy=2

then the equation is an exact since

dm/dx=dn/dy

then we integrate

∫mdy +∫ndx=c

∫(2x+3y-5)dy + (3x+2y-5)dx=c

⇒3y²/2-5y + 3x²/2-5x=c

Answer 2

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(2x+3y-5)dy/dx=-(3x+2y-5)

⇒ (2x+3y-5)dy=-(3x+2y-5)dx

then by compering with

mdy+ndx=0

⇒ m=(2x+3y-5) and n=(3x+2y-5)

dm/dx=2 and dn/dy=2

then the equation is an exact since

dm/dx=dn/dy

then we integrate

∫mdy +∫ndx=c

∫(2x+3y-5)dy + (3x+2y-5)dx=c

$=»3y²/2-5y + 3x²/2-5x=c$