φ_n :R→R n∈N^∗ φ_n =t^n (d^n φ/dt^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


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
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

	$ϕ_{2} (t) = t^{2} \frac{d^{2} ϕ_{2}}{{d t}^{2}}$ $t = e^{x}$ $\frac{d ϕ_{2}}{d t} = e^{- x} \frac{d ϕ_{2}}{d x}, \frac{d^{2} ϕ_{2}}{d t} = (\frac{d^{2} ϕ_{2}}{{d x}^{2}} - \frac{d ϕ_{2}}{d x}) e^{- 2 x}$ $ϕ_{2} = ϕ_{2}^{^{″}} - ϕ_{2}^{^{'}}$ $ϕ_{2}^{^{″}} - ϕ_{2}^{^{'}} - ϕ_{2} = 0$ $characteristic equation$ $r^{2} - r - 1 = 0$ $r_{1}, r_{2} = \frac{1 \pm \sqrt{5}}{2}$ $ϕ_{2} (x) = c_{1} e^{r_{1} x} + c_{2} e^{r_{2} x}$ $ϕ_{2} (t) = c_{1} t^{r_{1}} + c_{2} t^{r_{2}}$ $ϕ_{2} (t) = - 1 \Rightarrow c_{1} + c_{2} = - 1$ $ϕ_{2}^{'} = c_{1} r_{1} t^{r_{1} - 1} + c_{2} r_{2} t^{r_{2} - 1}$ $ϕ_{2}^{'} (1) = 1$ $1 = c_{1} r_{1} + c_{2} r_{2} = c_{1} (\frac{1 + \sqrt{5}}{2}) - c_{1} (\frac{1 - \sqrt{5}}{2}) - \frac{1 - \sqrt{5}}{2}$ $1 + \frac{1 - \sqrt{5}}{2} = c_{1} \sqrt{5} \Rightarrow c_{1} = \frac{3 - \sqrt{5}}{\sqrt{5}}$ $c_{2} = - 1 - c_{1} = - 1 - \frac{3 - \sqrt{5}}{\sqrt{5}} = \frac{2 \sqrt{5} - 3}{2}$ $ϕ_{2} (t) = c_{1} x^{r_{1}} + c_{2} x^{r_{2}}, r_{1} = \frac{1 + \sqrt{5}}{2}, r_{2} = \frac{1 - \sqrt{5}}{2}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum