Solve-y-t-tanh-y-t-1-y-0-2-tanh-is-hyperbolic-tangent-function-

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^′ =(1/2)ln(((1+t)/(1−t)))⇒y=(1/2)(ln(1−t^2 )+tln(((1+t)/(1−t))))+c y(0)=2⇒c=2 y=(1/2)(ln(1−t^2 )+tln(((1+t)/(1−t))))+2

$${y}^{'} =\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\right)\Rightarrow{y}=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+{tln}\left(\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\right)\right)+{c} \\ $$$${y}\left(\mathrm{0}\right)=\mathrm{2}\Rightarrow{c}=\mathrm{2} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+{tln}\left(\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\right)\right)+\mathrm{2} \\ $$$$ \\ $$

Commented by mathocean1 last updated on 26/Mar/23

thanks... but what happened to the exponent −1 of tanh...?

$${thanks}…\: \\ $$$${but}\:{what}\:{happened}\:{to}\:{the}\:{exponent} \\ $$$$−\mathrm{1}\:{of}\:{tanh}…? \\ $$

Commented by mehdee42 last updated on 27/Mar/23

sire : you mean (f)^(−1) revers function or (1/f) ?

$${sire}\::\:{you}\:{mean}\:\left({f}\right)^{−\mathrm{1}} \:{revers}\:{function}\:{or}\:\frac{\mathrm{1}}{{f}}\:\:? \\ $$

Commented by mehdee42 last updated on 27/Mar/23

if f(x)=tanhx=((e^(2x) −1)/(e^(2x) +1))=y⇒e^(2x) =((y+1)/(1−y))⇒f^(−1) (x)=(1/2)ln(((x+1)/(1−x)))

$${if}\:\:{f}\left({x}\right)={tanhx}=\frac{{e}^{\mathrm{2}{x}} −\mathrm{1}}{{e}^{\mathrm{2}{x}} +\mathrm{1}}={y}\Rightarrow{e}^{\mathrm{2}{x}} =\frac{{y}+\mathrm{1}}{\mathrm{1}−{y}}\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{x}+\mathrm{1}}{\mathrm{1}−{x}}\right) \\ $$

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^′ =(1/(tanhy))⇒tanhydy=dt⇒ln(coshy)=t+c y(0)=2⇒c=ln(cosh2)⇒ln(coshy)=t+ln(cosh2) ⇒coshy=e^t cosh2⇒y=cosh^(−1) (e^t cosh2) ⇒y=ln(e^t cosh2+(√((e^t cosh2)^2 −1)))

$${Ok} \\ $$$${y}^{'} =\frac{\mathrm{1}}{{tanhy}}\Rightarrow{tanhydy}={dt}\Rightarrow{ln}\left({coshy}\right)={t}+{c} \\ $$$${y}\left(\mathrm{0}\right)=\mathrm{2}\Rightarrow{c}={ln}\left({cosh}\mathrm{2}\right)\Rightarrow{ln}\left({coshy}\right)={t}+{ln}\left({cosh}\mathrm{2}\right) \\ $$$$\Rightarrow{coshy}={e}^{{t}} {cosh}\mathrm{2}\Rightarrow{y}={cosh}^{−\mathrm{1}} \left({e}^{{t}} {cosh}\mathrm{2}\right) \\ $$$$\Rightarrow{y}={ln}\left({e}^{{t}} {cosh}\mathrm{2}+\sqrt{\left({e}^{{t}} {cosh}\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}}\right) \\ $$

Commented by mathocean1 last updated on 29/Mar/23

$${Thank}\:{very}\:{much}\:{sir} \\ $$

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