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Question Number 1023 by 123456 last updated on 20/May/15

ϕ_{n} : R \to R

n \in N^{*}

ϕ_{n} = t^{n} \frac{d^{n} ϕ}{{d t}^{n}}

ϕ_{1} (t) = ?, ϕ_{1} (1) = + 1

ϕ_{2} (t) = ?, ϕ_{2} (1) = - 1, ϕ_{2}^{^{'}} (1) = + 1

ϕ_{3} (t) = ?, ϕ_{3} (1) = + 1, ϕ_{3}^{^{'}} (1) = - 1, ϕ_{3}^{^{″}} (1) = + 1

Commented by prakash jain last updated on 21/May/15

ϕ_{1} (t) = t \frac{d ϕ_{1} (t)}{d t} \Rightarrow \ln y = \ln t + C

ϕ_{1} (t) = k t

ϕ_{1} (1) = 1 \Rightarrow k = 1

ϕ_{1} (t) = t

Answered by prakash jain last updated on 21/May/15

ϕ_{2} (t) = t^{2} \frac{d^{2} ϕ_{2}}{{d t}^{2}}

t = e^{x}

\frac{d ϕ_{2}}{d t} = e^{- x} \frac{d ϕ_{2}}{d x}, \frac{d^{2} ϕ_{2}}{d t} = (\frac{d^{2} ϕ_{2}}{{d x}^{2}} - \frac{d ϕ_{2}}{d x}) e^{- 2 x}

ϕ_{2} = ϕ_{2}^{^{″}} - ϕ_{2}^{^{'}}

ϕ_{2}^{^{″}} - ϕ_{2}^{^{'}} - ϕ_{2} = 0

characteristic equation

r^{2} - r - 1 = 0

r_{1}, r_{2} = \frac{1 \pm \sqrt{5}}{2}

ϕ_{2} (x) = c_{1} e^{r_{1} x} + c_{2} e^{r_{2} x}

ϕ_{2} (t) = c_{1} t^{r_{1}} + c_{2} t^{r_{2}}

ϕ_{2} (t) = - 1 \Rightarrow c_{1} + c_{2} = - 1

ϕ_{2}^{'} = c_{1} r_{1} t^{r_{1} - 1} + c_{2} r_{2} t^{r_{2} - 1}

ϕ_{2}^{'} (1) = 1

1 = c_{1} r_{1} + c_{2} r_{2} = c_{1} (\frac{1 + \sqrt{5}}{2}) - c_{1} (\frac{1 - \sqrt{5}}{2}) - \frac{1 - \sqrt{5}}{2}

1 + \frac{1 - \sqrt{5}}{2} = c_{1} \sqrt{5} \Rightarrow c_{1} = \frac{3 - \sqrt{5}}{\sqrt{5}}

c_{2} = - 1 - c_{1} = - 1 - \frac{3 - \sqrt{5}}{\sqrt{5}} = \frac{2 \sqrt{5} - 3}{2}

ϕ_{2} (t) = c_{1} x^{r_{1}} + c_{2} x^{r_{2}}, r_{1} = \frac{1 + \sqrt{5}}{2}, r_{2} = \frac{1 - \sqrt{5}}{2}