Find D^2y/dx^2 In Terms Of X And Y For A) Xy+y^2 =

CrazyDan

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Find d^2y/dx^2 in terms of x and y for a) xy+y^2 =1 b) x^2 +xy = 5

 

Answer No.1

Anonymous

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A)d^2y/dx^2=2y/((x+2y)^2) b)d^2y/dx^2=(2y/X^2)+2(1/x)

Answer No.2

DrSupa9

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A) the equation will be xy+y^2-1=0 differentiating w.r.t x y + xdy/dx +2ydy/dx = 0 (x+2y)dy/dx = -y dy/dx = -y/(x+2y) b) the equation is x^2+xy-5=0 differentiating w.r .t x 2x + y + xdy/dx = 0 dy/dx = -(2x+y)/x

Answer No.3

Anonymous

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This will be solved implicitely a) the equation will be xy+y^2-1=0 differentiating w.r.t x y + xdy/dx +2ydy/dx = 0 (x+2y)dy/dx = -y dy/dx = -y/(x+2y) differentiating in the same way, we can get d^2y/dx^2 = y/(x+2y)^2 b) the equation is x^2+xy-5=0 differentiating w.r .t x 2x + y + xdy/dx = 0 dy/dx = -(2x+y)/x differentiating in the same way again d^2y/dx^2 = 2y/x^2 +2y/x

Answer No.4

Anonymous

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A. xy+y^2-1=0 differentiating w.r.t x y + xdy/dx +2ydy/dx = 0 (x+2y)dy/dx = -y dy/dx = -y/(x+2y) differentiating in the same way, we can get d^2y/dx^2 = y/(x+2y)^2 b) x^2+xy-5=0 differentiating w.r .t x 2x + y + xdy/dx = 0 dy/dx = -(2x+y)/x differentiating in the same way again d^2y/dx^2 = 2y/x^2 +2y/x

Answer No.5

Anonymous

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A) xy+y2-1=0 Differentiating w.r.t x: y + xdy/dx +2ydy/dx = 0 (x+2y)dy/dx = -y dy/dx = -y/(x+2y) b) x2+xy-5=0 Differentiating w.r .t x : 2x + y + xdy/dx = 0 ∴dy/dx = -(2x+y)/x

Answer No.6

Anonymous

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Xy - 2x = 3 xy = 2x +3 Differentiating both sides w.r.t. x, x. 2yy' + y. 2x = 2 x.yy' + y.x = 1................... (1) xy. y' = 1 - xy y' = ( 1 - xy ) / ( xy ) ........... (2) Now, differentiating (1) w.r.t. x again, [ xy. y'' + y' ( xy' + y.2x ) ] + [ y.1 + x.2yy' ] = 0 xy. y'' + x (y') + 2xy.y' + y + 2xy.y' = 0 - xy. y'' = x (y') + y + 4xy.y' = [ x (1-xy) / (xy) ] + y + [ 4xy. (1-xy) / (xy) ] = [ (1-xy) / (xy) ] + y + [ 4(1-xy) / x ] = [ 1/(xy) ] - (1/x) + y + (4/x) - 4y = [ 1/(xy) ] + (3/x) - 3y y'' = { [ 1/(xy) ] + (3/x) - 3y } / ( - xy )

Answer No.7

Anonymous

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Sol. as we know that xy - 2x = 3 xy = 2x +3 Differentiating both sides w.r.t. x.... x. 2yy' + y. 2x = 2 x.yy' + y.x = 1................... (1) xy. y' = 1 - xy y' = ( 1 - xy ) / ( xy ) ........... (2) Now, differentiating (1) w.r.t. x again, [ xy. y'' + y' ( xy' + y.2x ) ] + [ y.1 + x.2yy' ] = 0 xy. y'' + x (y') + 2xy.y' + y + 2xy.y' = 0 - xy. y'' = x (y') + y + 4xy.y' = [ x (1-xy) / (xy) ] + y + [ 4xy. (1-xy) / (xy) ] = [ (1-xy) / (xy) ] + y + [ 4(1-xy) / x ] = [ 1/(xy) ] - (1/x) + y + (4/x) - 4y = [ 1/(xy) ] + (3/x) - 3y y'' = { [ 1/(xy) ] + (3/x) - 3y } / ( - xy )

Answer No.8

Anonymous

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A.2y(x+y)/(x+2y)^3 b.2(x+y)/x^2

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