Solve : { ((y′(t)=[tanh(y(t))]^(−1) )),((y(0)=2)) :} tanh is hyperbolic tangent function.


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Question Number 190056 by mathocean1 last updated on 26/Mar/23
$Solve : { ((y′(t)=[tanh(y(t))]^(−1) )),((y(0)=2)) :} tanh is hyperbolic tangent function.$
$S o l v e :$ ${\begin{cases} y^{'} (t) = {[t a n h (y (t))]}^{- 1} \\ y (0) = 2 \end{cases}$ $t a n h i s h y p e r b o l i c t a n g e n t f u n c t i o n .$
	Answered by mehdee42 last updated on 26/Mar/23

	$y^{^{'}} = \frac{1}{2} l n (\frac{1 + t}{1 - t}) \Rightarrow y = \frac{1}{2} (l n (1 - t^{2}) + t l n (\frac{1 + t}{1 - t})) + c$ $y (0) = 2 \Rightarrow c = 2$ $y = \frac{1}{2} (l n (1 - t^{2}) + t l n (\frac{1 + t}{1 - t})) + 2$
	Commented by mathocean1 last updated on 26/Mar/23

	$t h a n k s . . .$ $b u t w h a t h a p p e n e d t o t h e e x p o n e n t$ $- 1 o f t a n h . . . ?$
	Commented by mehdee42 last updated on 27/Mar/23

	$s i r e : y o u m e a n {(f)}^{- 1} r e v e r s f u n c t i o n o r \frac{1}{f} ?$
	Commented by mehdee42 last updated on 27/Mar/23

	$i f f (x) = t a n h x = \frac{e^{2 x} - 1}{e^{2 x} + 1} = y \Rightarrow e^{2 x} = \frac{y + 1}{1 - y} \Rightarrow f^{- 1} (x) = \frac{1}{2} l n (\frac{x + 1}{1 - x})$
	Commented by mathocean1 last updated on 27/Mar/23

	$y e s i m e a n \frac{1}{f} . . .$
	Commented by mehdee42 last updated on 28/Mar/23

	$O k$ $y^{^{'}} = \frac{1}{t a n h y} \Rightarrow t a n h y d y = d t \Rightarrow l n (c o s h y) = t + c$ $y (0) = 2 \Rightarrow c = l n (c o s h 2) \Rightarrow l n (c o s h y) = t + l n (c o s h 2)$ $\Rightarrow c o s h y = e^{t} c o s h 2 \Rightarrow y = {c o s h}^{- 1} (e^{t} c o s h 2)$ $\Rightarrow y = l n (e^{t} c o s h 2 + \sqrt{{(e^{t} c o s h 2)}^{2} - 1})$
	Commented by mathocean1 last updated on 29/Mar/23

	$T h a n k v e r y m u c h s i r$
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