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Question Number 54880 by Otchere Abdullai last updated on 14/Feb/19

If x^2 −y^2 =a^2 find (d^2 y/dx^2 ) if a is constant.

$${If}\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \:\:{find}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:{if}\:{a}\:{is} \\ $$$${constant}. \\ $$

Answered by kaivan.ahmadi last updated on 14/Feb/19

f(x,y)=x^2 −y^2 −a^2 =0 y_x ^′ =(dy/dx)=−(f_x /f_y )=−((2x)/(−2y))=(x/y) (d^2 y/dx^2 )=((1×y−y_x ^′ ×x)/y^2 )=((y−(x^2 /y))/y^2 )=((y^2 −x^2 )/y^3 )

$$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{y}_{\mathrm{x}} ^{'} =\frac{\mathrm{dy}}{\mathrm{dx}}=−\frac{\mathrm{f}_{\mathrm{x}} }{\mathrm{f}_{\mathrm{y}} }=−\frac{\mathrm{2x}}{−\mathrm{2y}}=\frac{\mathrm{x}}{\mathrm{y}} \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }=\frac{\mathrm{1}×\mathrm{y}−\mathrm{y}_{\mathrm{x}} ^{'} ×\mathrm{x}}{\mathrm{y}^{\mathrm{2}} }=\frac{\mathrm{y}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{y}}}{\mathrm{y}^{\mathrm{2}} }=\frac{\mathrm{y}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }{\mathrm{y}^{\mathrm{3}} } \\ $$

Commented by Otchere Abdullai last updated on 14/Feb/19

Thanks sir!

$${Thanks}\:{sir}! \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19

y^2 =x^2 −a^2 2y(dy/dx)=2x→(dy/dx)=(x/y) y×(d^2 y/dx^2 )+((dy/dx))^2 =1 y((d^2 y/dx^2 ))=1−((x/y))^2 (d^2 y/dx^2 )=((y^2 −x^2 )/y^3 )→(d^2 y/dx^2 )=((−a^2 )/y^3 ) (d^2 y/dx^2 )=((−a^2 )/((x^2 −a^2 )^(3/2) ))

$${y}^{\mathrm{2}} ={x}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$\mathrm{2}{y}\frac{{dy}}{{dx}}=\mathrm{2}{x}\rightarrow\frac{{dy}}{{dx}}=\frac{{x}}{{y}} \\ $$$${y}×\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${y}\left(\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)=\mathrm{1}−\left(\frac{{x}}{{y}}\right)^{\mathrm{2}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} }{{y}^{\mathrm{3}} }\rightarrow\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{−{a}^{\mathrm{2}} }{{y}^{\mathrm{3}} } \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{−{a}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$ \\ $$

Commented by Otchere Abdullai last updated on 14/Feb/19

Thanks prof

$${Thanks}\:{prof} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19

i am not a professor but 49years old indian interestd in mathematics physics and mythology ,metaphysics unsolved mystry...

$${i}\:{am}\:{not}\:{a}\:{professor}\:{but}\:\mathrm{49}{years}\:{old}\:{indian}\:{interestd} \\ $$$${in}\:{mathematics}\:{physics}\:{and}\:{mythology}\:,{metaphysics} \\ $$$${unsolved}\:{mystry}... \\ $$

Commented by Otchere Abdullai last updated on 14/Feb/19

yes i gave you the professor because you have done great contribution to the group

$${yes}\:{i}\:{gave}\:{you}\:{the}\:{professor}\:{because} \\ $$$${you}\:{have}\:{done}\:{great}\:{contribution}\:{to} \\ $$$${the}\:{group} \\ $$

Commented by $@ty@m last updated on 14/Feb/19

Yeah. He desereves the title of “Honorary Professor”.

$${Yeah}. \\ $$$${He}\:{desereves}\:{the}\:{title}\:{of}\:``{Honorary}\:{Professor}''. \\ $$

Commented by Otchere Abdullai last updated on 14/Feb/19

Yes you are right don′t forget Mr W was the first honoured professor so i always use prof W instered of Mr.

$${Yes}\:{you}\:{are}\:{right}\:{don}'{t}\:{forget}\:{Mr}\:{W} \\ $$$${was}\:{the}\:{first}\:{honoured}\:{professor} \\ $$$${so}\:{i}\:{always}\:{use}\:{prof}\:{W}\:{instered}\:{of}\:{Mr}. \\ $$

Commented by rahul 19 last updated on 14/Feb/19

Not only him , but MrW sir, Mjs sir, Ajfour sir toooo!

$${Not}\:{only}\:{him}\:,\:{but}\:{MrW}\:{sir},\:{Mjs}\:{sir}, \\ $$$${Ajfour}\:{sir}\:{toooo}! \\ $$

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