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Show that \(\operatorname{erc}(x)\) satisfies the differential equation
\[\frac{\mathrm dy}{\mathrm dx}=2xy\frac2{\sqrt \pi}\] With \(y=1\) when \(x=0\)
 one year ago
 one year ago
Show that \(\operatorname{erc}(x)\) satisfies the differential equation \[\frac{\mathrm dy}{\mathrm dx}=2xy\frac2{\sqrt \pi}\] With \(y=1\) when \(x=0\)
 one year ago
 one year ago

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UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[ \newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\pa[2]{\frac{\partial #1}{\partial #2}} % partial derivative \newcommand\erf[1]{\operatorname {erf}\left(#1\right)} % Error function erf(x) \newcommand\erfint[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{u^2}}{u}} % Error function integral erfc(x) \newcommand\erfc[1]{\operatorname {erfc}\left(#1\right)} % Complimentary Error function Compliment (x) \newcommand\erfcint[1]{\frac2{\sqrt \pi}\intl{#1}{\infty}{e^{u^2}}u} % Complimentary Error function integral (x) \newcommand\erc[1]{\operatorname{erc}\left(#1\right)} % Normalised Error function (x) \newcommand\ercint[1]{e^{#1^2}\erfcint{#1}} % Normalised Error function integral (x) \begin{equation*} \de yx=2xy\frac2{\sqrt \pi},\qquad\qquad y(0)=1\\ \end{equation*}\]\[ \begin{align*} y(x)&=\erc x\\ &=e^{x^2}\erfc x\\ &=\ercint x\\ \\ y(0)&=\erfcint 0\\ &=1\\ \\ \de yx&=\frac2{\sqrt \pi}\left[\de{\left(e^{x^2}\right)}x\intl x\infty{e^{u^2}}u+e^{x^2}\intl x\infty {\pa {\left(e^{u^2}\right)}u}u\right]\\ &=\frac2{\sqrt \pi}\left[2xe^{x^2}\intl x\infty{e^{u^2}}u+e^{x^2}\left.e^{u^2}\right_x^\infty\right]\\ &=\frac2{\sqrt \pi}\left[2xe^{x^2}\intl x\infty{e^{u^2}}ue^{x^2}e^{x^2}\right]\\ &=2xe^{x^2}\frac{2}{\sqrt \pi}\intl x\infty{e^{u^2}}u\frac2{\sqrt \pi}\\ &=2xy\frac2{\sqrt \pi}\\ \end{align*} \]
 one year ago

Goten77Best ResponseYou've already chosen the best response.0
im sorry but man... im good at math but i only know stuff up to calculus 2.... anything after i havent taken yet.... best of luck
 one year ago

chihiroasleafBest ResponseYou've already chosen the best response.0
what is erc(x) ?
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
my book defines:: ___ The error function:\[\operatorname {erf}(x)=\frac{2}{\sqrt \pi}\int\limits_0^xe^{u^2}\mathrm du\]___ The complementary error function:\[\operatorname{erfc}(x)=1\operatorname {erf}(x)\] ___\[\operatorname {erc}(x)=e^{x^2}\operatorname{erfc(x)}\]
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
i guess it is the normalised complementary error function
 one year ago

chihiroasleafBest ResponseYou've already chosen the best response.0
ahh I see..... what's your problem? I think you've done it..
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
so i haven't made any errors?
 one year ago

chihiroasleafBest ResponseYou've already chosen the best response.0
It seems correct..
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
\[\Large {y=\operatorname{erc}(x)=e^{x^2}\left[1\frac{2}{\sqrt \pi}\int_0^xe^{u^2}\;\mathrm du\right]\\\frac{dy}{dx}=2xe^{x^2}\left[1\frac{2}{\sqrt \pi}\int_0^xe^{u^2}\;\mathrm du\right]+e^{x^2}\left[\frac{2}{\sqrt \pi}e^{x^2}\right]\\=2xy\frac{2}{\sqrt \pi}}\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
it's shorter if you apply the fundamental theorem of calculus
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
you used the derivative "under the integral sign", and i appreciate looking at a different solution. i see the function \(\operatorname{efc}(x)\) as a product, so i sued the product rule for derivative. in the process, i also saw the fundamental theorem of calculus\[\frac{\mathrm d}{\mathrm dx} \int_a^x f(u)\;\mathrm du=f(x)\]
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
the only difference i can see your method is that you used erf where i used erfc
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
I can't say I understand the difference between derivative "under the integral sign" & the fundamental theorem of calculus
 one year ago
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