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Question Number 8115 by uchechukwu okorie favour last updated on 30/Sep/16
if xy+y^2 =1. Find (d^2 y/dx) at (0,1)
$${if}\:{xy}+{y}^{\mathrm{2}} =\mathrm{1}.\:{Find}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}}\:{at}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$
Answered by prakash jain last updated on 30/Sep/16
x(dy/dx)+y+2y(dy/dx)=0...(i) x(d^2 y/dx^2 )+(dy/dx)+(dy/dx)+2y(d^2 y/dx^2 )+2((dy/dx))^2 ...(ii) (x+2y)(d^2 y/dx^2 )=−2((dy/dx)+((dy/dx)))^2 from (i) (dy/dx)=−(y/(x+2y)) (d^2 y/dx^2 )=((−2(−(y/(x+2y))+(−(y/(x+2y)))^2 ))/((x+2y))) x=0,y=1 (d^2 y/dx^2 )=((−2(−(1/2)+(1/4)))/2)=−1(−(1/4))=(1/4)
$${x}\frac{{dy}}{{dx}}+{y}+\mathrm{2}{y}\frac{{dy}}{{dx}}=\mathrm{0}…\left({i}\right) \\ $$$${x}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\frac{{dy}}{{dx}}+\frac{{dy}}{{dx}}+\mathrm{2}{y}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{2}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:\:…\left({ii}\right) \\ $$$$\left({x}+\mathrm{2}{y}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=−\mathrm{2}\left(\frac{{dy}}{{dx}}+\left(\frac{{dy}}{{dx}}\right)\right)^{\mathrm{2}} \\ $$$${from}\:\left({i}\right)\:\:\frac{{dy}}{{dx}}=−\frac{{y}}{{x}+\mathrm{2}{y}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{−\mathrm{2}\left(−\frac{{y}}{{x}+\mathrm{2}{y}}+\left(−\frac{{y}}{{x}+\mathrm{2}{y}}\right)^{\mathrm{2}} \right)}{\left({x}+\mathrm{2}{y}\right)} \\ $$$${x}=\mathrm{0},{y}=\mathrm{1} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{−\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{2}}=−\mathrm{1}\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Answered by mrW1 last updated on 13/Feb/17
x=((1−y^2 )/y)=(1/y)−y (dx/dy)=−(1/y^2 )−1=−((1+y^2 )/y^2 ) g=(dy/dx)=(1/(dx/dy))=−(y^2 /(1+y^2 ))=−1+(1/(1+y^2 )) (dg/dy)=−((2y)/((1+y^2 )^2 )) (d^2 y/dx^2 )=(dg/dx)=(dg/dy)×(dy/dx)=((2y^3 )/((1+y^2 )^3 ))=2((y/(1+y^2 )))^3 with y=1, (d^2 y/dx^2 )=(1/4)
$${x}=\frac{\mathrm{1}−{y}^{\mathrm{2}} }{{y}}=\frac{\mathrm{1}}{{y}}−{y} \\ $$$$\frac{{dx}}{{dy}}=−\frac{\mathrm{1}}{{y}^{\mathrm{2}} }−\mathrm{1}=−\frac{\mathrm{1}+{y}^{\mathrm{2}} }{{y}^{\mathrm{2}} } \\ $$$${g}=\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\frac{{dx}}{{dy}}}=−\frac{{y}^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }=−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} } \\ $$$$\frac{{dg}}{{dy}}=−\frac{\mathrm{2}{y}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{dg}}{{dx}}=\frac{{dg}}{{dy}}×\frac{{dy}}{{dx}}=\frac{\mathrm{2}{y}^{\mathrm{3}} }{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)^{\mathrm{3}} }=\mathrm{2}\left(\frac{{y}}{\mathrm{1}+{y}^{\mathrm{2}} }\right)^{\mathrm{3}} \\ $$$${with}\:{y}=\mathrm{1},\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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