3x+xy+4y=8
3 + y + xy' + 4y' = 0
y'(x+4) = -(y+3)
y' = -(y+3)/(x+4)
y" = -[(y')(x+4)-(y+3)(1)]/(x+4)^2
= -[-(y+3)/(x+4)*(x+4) - (y+3)]/(x+4)^2
= -(-3-y-y-3)/(x+4)^2
= (2y+6)/(x+4)^2
check:
3x+xy+4y=8
y = (8-3x)/(x+4)
y' = -20/(x+4)^2
y" = 40/(x+4)^3
Since y = (8-3x)/(x+4), that means
2y+6 = (16-6x + 6(x+4))/(x+4)
= (16-6x+6x+24)/(x+4)
= 40/(x+4)
So,
y" = (2y+6)/(x+4)^2
= 40/(x+4)^3
So the implicit and explicit vales agree.
If 3x+xy+4y=8, what is the value of d^2y/dx^2 at the point (1,1)? My work:
3+(x)(dy/dx)+(y)+4(dy/dx)=0
dy/dx(x+4)=-3-y
dy/dx=-3-y/x+4
2nd Dervative: (x+4)(0-dy/dx)-(-3-y)/(x+4)^2
(x+4)(-(-y-3/x+4))-(-3-y)/(x+4)^2
(x+4)(y+3)/(-x+4)+(y+3)/(x+4)^2
*Mult. everything by (-x+4)
Simplify:
(x+4)(y+3)+(y+3)(-x+4)/(x+4)^2(-x+4)
*The (-x+4) cancels
Left over: (x+4)(y+3)+(y+3)/(x+4)^2
*Plug in (1,1)
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