Question-204418

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{d x}{d t} = e^{t} = x

\frac{d y}{d x} = \frac{d y}{d t} \times \frac{1}{\frac{d x}{d t}} = \frac{d y}{d t} \times \frac{1}{x}

\Rightarrow x \frac{d y}{d x} = \frac{d y}{d t}

\frac{d^{2} y}{{d x}^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d t} (\frac{d y}{d x}) \times \frac{1}{\frac{d x}{d t}} = \frac{d}{d t} (\frac{d y}{d t} \times \frac{1}{x}) \times \frac{1}{x}

= (\frac{d^{2} y}{{d t}^{2}} \times \frac{1}{x} + \frac{d y}{d t} \times (- \frac{1}{x^{2}} \times \frac{d x}{d t})) \times \frac{1}{x}

= (\frac{d^{2} y}{{d t}^{2}} \times \frac{1}{x} + \frac{d y}{d t} \times (- \frac{1}{x^{2}} \times x)) \times \frac{1}{x}

= (\frac{d^{2} y}{{d t}^{2}} - \frac{d y}{d t}) \times \frac{1}{x^{2}}

\Rightarrow x^{2} \frac{d^{2} y}{{d x}^{2}} = \frac{d^{2} y}{{d t}^{2}} - \frac{d y}{d t}

x^{2} \frac{d^{2} y}{{d x}^{2}} + x \frac{d y}{d x} - 4 y = 16

\frac{d^{2} y}{{d t}^{2}} - \frac{d y}{d t} + \frac{d y}{d t} - 4 y = 16

\frac{d^{2} y}{{d t}^{2}} - 4 y = 16

\frac{d^{2} (y + 4)}{{d t}^{2}} - 4 (y + 4) = 0

\Rightarrow y + 4 = A e^{2 t} + B e^{- 2 t}

y (x = 1) = 0 \Rightarrow y (t = 0) = 0 \Rightarrow

0 + 4 = A + B

\frac{d y}{d x} ∣_{x = 1} = 0 \Rightarrow \frac{d y}{d t} ∣_{t = 0} = 0 \Rightarrow

0 = 2 A - 2 B

\Rightarrow A = B = 2

\Rightarrow y + 4 = 2 (e^{2 t} + e^{- 2 t}) = 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

thank you