CrazyDan From: US Posts: 41 Votes: 164 |
Find d^2y/dx^2 in terms of x and y for a) xy+y^2 =1 b) x^2 +xy = 5 |

Anonymous From: - Posts: - Votes: - |
A)d^2y/dx^2=2y/((x+2y)^2) b)d^2y/dx^2=(2y/X^2)+2(1/x) |

DrSupa9 From: US Posts: 36 Votes: 144 |
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 |

Anonymous From: - Posts: - Votes: - |
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 |

Anonymous From: - Posts: - Votes: - |
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 |

Anonymous From: - Posts: - Votes: - |
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 |

Anonymous From: - Posts: - Votes: - |
X²·y² - 2x = 3 x²·y² = 2x +3 Differentiating both sides w.r.t. x, x². 2yy' + y². 2x = 2 x².yy' + y².x = 1................... (1) x²y. y' = 1 - xy² y' = ( 1 - xy² ) / ( x²y ) ........... (2) Now, differentiating (1) w.r.t. x again, [ x²y. y'' + y' ( x²y' + y.2x ) ] + [ y².1 + x.2yy' ] = 0 x²y. y'' + x² (y')² + 2xy.y' + y² + 2xy.y' = 0 - x²y. y'' = x² (y')² + y² + 4xy.y' = [ x² (1-xy²)² / (x²y)² ] + y² + [ 4xy. (1-xy²) / (x²y) ] = [ (1-xy²) / (x²y²) ] + y² + [ 4(1-xy²) / x ] = [ 1/(x²y²) ] - (1/x) + y² + (4/x) - 4y² = [ 1/(x²y²) ] + (3/x) - 3y² y'' = { [ 1/(x²y²) ] + (3/x) - 3y² } / ( - x²y ) |

Anonymous From: - Posts: - Votes: - |
Sol. as we know that x²·y² - 2x = 3 x²·y² = 2x +3 Differentiating both sides w.r.t. x.... x². 2yy' + y². 2x = 2 x².yy' + y².x = 1................... (1) x²y. y' = 1 - xy² y' = ( 1 - xy² ) / ( x²y ) ........... (2) Now, differentiating (1) w.r.t. x again, [ x²y. y'' + y' ( x²y' + y.2x ) ] + [ y².1 + x.2yy' ] = 0 x²y. y'' + x² (y')² + 2xy.y' + y² + 2xy.y' = 0 - x²y. y'' = x² (y')² + y² + 4xy.y' = [ x² (1-xy²)² / (x²y)² ] + y² + [ 4xy. (1-xy²) / (x²y) ] = [ (1-xy²) / (x²y²) ] + y² + [ 4(1-xy²) / x ] = [ 1/(x²y²) ] - (1/x) + y² + (4/x) - 4y² = [ 1/(x²y²) ] + (3/x) - 3y² y'' = { [ 1/(x²y²) ] + (3/x) - 3y² } / ( - x²y ) |

Anonymous From: - Posts: - Votes: - |
A.2y(x+y)/(x+2y)^3 b.2(x+y)/x^2 |

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