Menu Close

Question-206442

Tinku Tara 0 Comments
Pin




Question Number 206442 by universe last updated on 14/Apr/24
Answered by Berbere last updated on 14/Apr/24
let f(x)=e^(g(x)) ; particular Solution just to simplifie the problems;f(0)=1=e^(g(0)) ⇒g(0)=0 ⇒f′(x)=g′e^(g(x)) ;f′′(x)=(g′^2 +g′′)e^(g(x)) ⇒(g′^2 +g′′)=x^2 −1;g′=t t^2 +t′=x^2 −1;(t=−x worck g(x)=−(x^2 /2) let f(x)=k(x)e^(−(x^2 /2)) ⇒f′(x)=(−xk+k′)e^(−(x^2 /2)) f′′(x)=(−k−xk′+k′′+x^2 k−xk′) Eq ⇔(k′′−2xk′+x^2 k−k)e^(−(x^2 /2)) =(x^2 −1)ke^(−(x^2 /2)) ⇔k′′−2xk′=0;h(x)=k′ ⇒h′(x)−2xh(x)=0⇒h(x)=ke^x^2 k(x)=∫ke^x^2 dx=c+k∫e^x^2 dx ∫_0 ^t e^x^2 =(2/( (√π)))erfi(t) k(x)=c+(2/( (√π)))k.erfi(x) f(x)=((2k)/( (√π)))erfi(x)e^(−(x^2 /2)) +ce^(−(x^2 /2)) f(0)=1⇒c=1;erfi(0)=0; f(x)=((2.a)/( (√π)))erfi(x)e^(−(x^2 /2)) +e^(−(x^2 /2)) =(c.erfi(x)+1)e^(−(x^2 /2)) ;c∈R
$${let}\:{f}\left({x}\right)={e}^{{g}\left({x}\right)} ;\:{particular}\:{Solution}\:{just}\:{to}\:{simplifie} \\ $$$${the}\:{problems};{f}\left(\mathrm{0}\right)=\mathrm{1}={e}^{{g}\left(\mathrm{0}\right)} \Rightarrow{g}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\Rightarrow{f}'\left({x}\right)={g}'{e}^{{g}\left({x}\right)} ;{f}''\left({x}\right)=\left({g}'^{\mathrm{2}} +{g}''\right){e}^{{g}\left({x}\right)} \\ $$$$\Rightarrow\left({g}'^{\mathrm{2}} +{g}''\right)={x}^{\mathrm{2}} −\mathrm{1};{g}'={t} \\ $$$${t}^{\mathrm{2}} +{t}'={x}^{\mathrm{2}} −\mathrm{1};\left({t}=−{x}\:{worck}\right. \\ $$$${g}\left({x}\right)=−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\: \\ $$$${let}\:{f}\left({x}\right)={k}\left({x}\right){e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \Rightarrow{f}'\left({x}\right)=\left(−{xk}+{k}'\right){e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$${f}''\left({x}\right)=\left(−{k}−{xk}'+{k}''+{x}^{\mathrm{2}} {k}−{xk}'\right) \\ $$$${Eq}\:\Leftrightarrow\left({k}''−\mathrm{2}{xk}'+{x}^{\mathrm{2}} {k}−{k}\right){e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} =\left({x}^{\mathrm{2}} −\mathrm{1}\right){ke}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\Leftrightarrow{k}''−\mathrm{2}{xk}'=\mathrm{0};{h}\left({x}\right)={k}' \\ $$$$\Rightarrow{h}'\left({x}\right)−\mathrm{2}{xh}\left({x}\right)=\mathrm{0}\Rightarrow{h}\left({x}\right)={ke}^{{x}^{\mathrm{2}} } \\ $$$${k}\left({x}\right)=\int{ke}^{{x}^{\mathrm{2}} } {dx}={c}+{k}\int{e}^{{x}^{\mathrm{2}} } {dx} \\ $$$$\int_{\mathrm{0}} ^{{t}} {e}^{{x}^{\mathrm{2}} } =\frac{\mathrm{2}}{\:\sqrt{\pi}}{erfi}\left({t}\right) \\ $$$${k}\left({x}\right)={c}+\frac{\mathrm{2}}{\:\sqrt{\pi}}{k}.{erfi}\left({x}\right) \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}{k}}{\:\sqrt{\pi}}{erfi}\left({x}\right){e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} +{ce}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1}\Rightarrow{c}=\mathrm{1};{erfi}\left(\mathrm{0}\right)=\mathrm{0}; \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}.{a}}{\:\sqrt{\pi}}{erfi}\left({x}\right){e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} +{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} =\left({c}.{erfi}\left({x}\right)+\mathrm{1}\right){e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} ;{c}\in\mathbb{R} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *