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Question-204418

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Question Number 204418 by peter frank last updated on 17/Feb/24
Commented by mr W last updated on 19/Feb/24
y=2(x−(1/x))^2
$${y}=\mathrm{2}\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \\ $$
Answered by mr W last updated on 19/Feb/24
x=e^t (dx/dt)=e^t =x (dy/dx)=(dy/dt)×(1/(dx/dt))=(dy/dt)×(1/x) ⇒x(dy/dx)=(dy/dt) (d^2 y/dx^2 )=(d/dx)((dy/dx))=(d/dt)((dy/dx))×(1/(dx/dt))=(d/dt)((dy/dt)×(1/x))×(1/x) =((d^2 y/dt^2 )×(1/x)+(dy/dt)×(−(1/x^2 )×(dx/dt)))×(1/x) =((d^2 y/dt^2 )×(1/x)+(dy/dt)×(−(1/x^2 )×x))×(1/x) =((d^2 y/dt^2 )−(dy/dt))×(1/x^2 ) ⇒x^2 (d^2 y/dx^2 )=(d^2 y/dt^2 )−(dy/dt) x^2 (d^2 y/dx^2 )+x(dy/dx)−4y=16 (d^2 y/dt^2 )−(dy/dt)+(dy/dt)−4y=16 (d^2 y/dt^2 )−4y=16 ((d^2 (y+4))/dt^2 )−4(y+4)=0 ⇒y+4=A e^(2t) +B e^(−2t) y(x=1)=0 ⇒y(t=0)=0 ⇒ 0+4=A+B (dy/dx)∣_(x=1) =0 ⇒(dy/dt)∣_(t=0) =0 ⇒ 0=2A−2B ⇒A=B=2 ⇒y+4=2(e^(2t) +e^(−2t) )=2(x^2 +(1/x^2 )) ⇒y=2(x^2 +(1/x^2 ))−4=2(x−(1/x))^2
$${x}={e}^{{t}} \\ $$$$\frac{{dx}}{{dt}}={e}^{{t}} ={x} \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{dt}}×\frac{\mathrm{1}}{\frac{{dx}}{{dt}}}=\frac{{dy}}{{dt}}×\frac{\mathrm{1}}{{x}} \\ $$$$\Rightarrow{x}\frac{{dy}}{{dx}}=\frac{{dy}}{{dt}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right)=\frac{{d}}{{dt}}\left(\frac{{dy}}{{dx}}\right)×\frac{\mathrm{1}}{\frac{{dx}}{{dt}}}=\frac{{d}}{{dt}}\left(\frac{{dy}}{{dt}}×\frac{\mathrm{1}}{{x}}\right)×\frac{\mathrm{1}}{{x}} \\ $$$$=\left(\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }×\frac{\mathrm{1}}{{x}}+\frac{{dy}}{{dt}}×\left(−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }×\frac{{dx}}{{dt}}\right)\right)×\frac{\mathrm{1}}{{x}} \\ $$$$=\left(\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }×\frac{\mathrm{1}}{{x}}+\frac{{dy}}{{dt}}×\left(−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }×{x}\right)\right)×\frac{\mathrm{1}}{{x}} \\ $$$$=\left(\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\frac{{dy}}{{dt}}\right)×\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\frac{{dy}}{{dt}} \\ $$$${x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{x}\frac{{dy}}{{dx}}−\mathrm{4}{y}=\mathrm{16} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\frac{{dy}}{{dt}}+\frac{{dy}}{{dt}}−\mathrm{4}{y}=\mathrm{16} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−\mathrm{4}{y}=\mathrm{16} \\ $$$$\frac{{d}^{\mathrm{2}} \left({y}+\mathrm{4}\right)}{{dt}^{\mathrm{2}} }−\mathrm{4}\left({y}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow{y}+\mathrm{4}={A}\:{e}^{\mathrm{2}{t}} +{B}\:{e}^{−\mathrm{2}{t}} \\ $$$${y}\left({x}=\mathrm{1}\right)=\mathrm{0}\:\Rightarrow{y}\left({t}=\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{0}+\mathrm{4}={A}+{B} \\ $$$$\frac{{dy}}{{dx}}\mid_{{x}=\mathrm{1}} =\mathrm{0}\:\Rightarrow\frac{{dy}}{{dt}}\mid_{{t}=\mathrm{0}} =\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{0}=\mathrm{2}{A}−\mathrm{2}{B} \\ $$$$\Rightarrow{A}={B}=\mathrm{2} \\ $$$$\Rightarrow{y}+\mathrm{4}=\mathrm{2}\left({e}^{\mathrm{2}{t}} +{e}^{−\mathrm{2}{t}} \right)=\mathrm{2}\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{y}=\mathrm{2}\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)−\mathrm{4}=\mathrm{2}\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \\ $$
Commented by peter frank last updated on 21/Feb/24
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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