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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

$$y=2{(x-\frac{1}{x})}^2$$

Answered by mr W last updated on 19/Feb/24

$$x=e^t$$$$\frac{dx}{dt}=e^t=x$$$$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{1}{\frac{dx}{dt}}=\frac{dy}{dt}\times \frac{1}{x}$$$$\Rightarrow x\frac{dy}{dx}=\frac{dy}{dt}$$$$\frac{d^{2}y}{{dx}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\times \frac{1}{\frac{dx}{dt}}=\frac{d}{dt}\left(\frac{dy}{dt}\times \frac{1}{x}\right)\times \frac{1}{x}$$$$=(\frac{d^{2}y}{{dt}^2}\times \frac{1}{x}+\frac{dy}{dt}\times (-\frac{1}{x^2}\times \frac{dx}{dt}))\times \frac{1}{x}$$$$=(\frac{d^{2}y}{{dt}^2}\times \frac{1}{x}+\frac{dy}{dt}\times (-\frac{1}{x^2}\times x))\times \frac{1}{x}$$$$=(\frac{d^{2}y}{{dt}^2}-\frac{dy}{dt})\times \frac{1}{x^2}$$$$\Rightarrow x^2\frac{d^{2}y}{{dx}^2}=\frac{d^{2}y}{{dt}^2}-\frac{dy}{dt}$$$$x^2\frac{d^{2}y}{{dx}^2}+x\frac{dy}{dx}-4y=16$$$$\frac{d^{2}y}{{dt}^2}-\frac{dy}{dt}+\frac{dy}{dt}-4y=16$$$$\frac{d^{2}y}{{dt}^2}-4y=16$$$$\frac{d^2(y+4)}{{dt}^2}-4(y+4)=0$$$$\Rightarrow y+4=A{e}^{2t}+B{e}^{-2t}$$$$y(x=1)=0\Rightarrow y(t=0)=0\Rightarrow$$$$0+4=A+B$$$$\frac{dy}{dx}{\mid }_{x=1}=0\Rightarrow \frac{dy}{dt}{\mid }_{t=0}=0\Rightarrow$$$$0=2A-2B$$$$\Rightarrow A=B=2$$$$\Rightarrow y+4=2(e^{2t}+e^{-2t})=2(x^2+\frac{1}{x^2})$$$$\Rightarrow y=2(x^2+\frac{1}{x^2})-4=2{(x-\frac{1}{x})}^2$$

Commented by peter frank last updated on 21/Feb/24

$$\mathrm{thank}\mathrm{you}$$

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