Question Number 190056 by mathocean1 last updated on 26/Mar/23
$${Solve}\:: \\ $$$$\begin{cases}{{y}'\left({t}\right)=\left[{tanh}\left({y}\left({t}\right)\right)\right]^{−\mathrm{1}} }\\{{y}\left(\mathrm{0}\right)=\mathrm{2}}\end{cases} \\ $$$$ \\ $$$${tanh}\:{is}\:{hyperbolic}\:{tangent}\:{function}. \\ $$
Answered by mehdee42 last updated on 26/Mar/23
$${y}^{'} =\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\right)\Rightarrow{y}=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+{tln}\left(\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\right)\right)+{c} \\ $$$${y}\left(\mathrm{0}\right)=\mathrm{2}\Rightarrow{c}=\mathrm{2} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+{tln}\left(\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\right)\right)+\mathrm{2} \\ $$$$ \\ $$
Commented by mathocean1 last updated on 26/Mar/23
$${thanks}…\: \\ $$$${but}\:{what}\:{happened}\:{to}\:{the}\:{exponent} \\ $$$$−\mathrm{1}\:{of}\:{tanh}…? \\ $$
Commented by mehdee42 last updated on 27/Mar/23
$${sire}\::\:{you}\:{mean}\:\left({f}\right)^{−\mathrm{1}} \:{revers}\:{function}\:{or}\:\frac{\mathrm{1}}{{f}}\:\:? \\ $$
Commented by mehdee42 last updated on 27/Mar/23
$${if}\:\:{f}\left({x}\right)={tanhx}=\frac{{e}^{\mathrm{2}{x}} −\mathrm{1}}{{e}^{\mathrm{2}{x}} +\mathrm{1}}={y}\Rightarrow{e}^{\mathrm{2}{x}} =\frac{{y}+\mathrm{1}}{\mathrm{1}−{y}}\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{x}+\mathrm{1}}{\mathrm{1}−{x}}\right) \\ $$
Commented by mathocean1 last updated on 27/Mar/23
$${yes}\:{i}\:{mean}\:\frac{\mathrm{1}}{{f}}… \\ $$
Commented by mehdee42 last updated on 28/Mar/23
$${Ok} \\ $$$${y}^{'} =\frac{\mathrm{1}}{{tanhy}}\Rightarrow{tanhydy}={dt}\Rightarrow{ln}\left({coshy}\right)={t}+{c} \\ $$$${y}\left(\mathrm{0}\right)=\mathrm{2}\Rightarrow{c}={ln}\left({cosh}\mathrm{2}\right)\Rightarrow{ln}\left({coshy}\right)={t}+{ln}\left({cosh}\mathrm{2}\right) \\ $$$$\Rightarrow{coshy}={e}^{{t}} {cosh}\mathrm{2}\Rightarrow{y}={cosh}^{−\mathrm{1}} \left({e}^{{t}} {cosh}\mathrm{2}\right) \\ $$$$\Rightarrow{y}={ln}\left({e}^{{t}} {cosh}\mathrm{2}+\sqrt{\left({e}^{{t}} {cosh}\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}}\right) \\ $$
Commented by mathocean1 last updated on 29/Mar/23
$${Thank}\:{very}\:{much}\:{sir} \\ $$