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UnkleRhaukus

  • 3 years ago

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

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  1. UnkleRhaukus
    • 3 years ago
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    \[ \newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\pa[2]{\frac{\partial #1}{\partial #2}} % partial derivative \newcommand\erf[1]{\operatorname {erf}\left(#1\right)} % Error function erf(x) \newcommand\erfint[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{-u^2}}{u}} % Error function integral erfc(x) \newcommand\erfc[1]{\operatorname {erfc}\left(#1\right)} % Complimentary Error function Compliment (x) \newcommand\erfcint[1]{\frac2{\sqrt \pi}\intl{#1}{\infty}{e^{-u^2}}u} % Complimentary Error function integral (x) \newcommand\erc[1]{\operatorname{erc}\left(#1\right)} % Normalised Error function (x) \newcommand\ercint[1]{e^{#1^2}\erfcint{#1}} % Normalised Error function integral (x) \begin{equation*} \de yx=2xy-\frac2{\sqrt \pi},\qquad\qquad y(0)=1\\ \end{equation*}\]\[ \begin{align*} y(x)&=\erc x\\ &=e^{x^2}\erfc x\\ &=\ercint x\\ \\ y(0)&=\erfcint 0\\ &=1\\ \\ \de yx&=\frac2{\sqrt \pi}\left[\de{\left(e^{x^2}\right)}x\intl x\infty{e^{-u^2}}u+e^{x^2}\intl x\infty {\pa {\left(e^{-u^2}\right)}u}u\right]\\ &=\frac2{\sqrt \pi}\left[2xe^{x^2}\intl x\infty{e^{-u^2}}u+e^{x^2}\left.e^{-u^2}\right|_x^\infty\right]\\ &=\frac2{\sqrt \pi}\left[2xe^{x^2}\intl x\infty{e^{-u^2}}u-e^{x^2}e^{-x^2}\right]\\ &=2xe^{x^2}\frac{2}{\sqrt \pi}\intl x\infty{e^{-u^2}}u-\frac2{\sqrt \pi}\\ &=2xy-\frac2{\sqrt \pi}\\ \end{align*} \]

  2. UnkleRhaukus
    • 3 years ago
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    y/n ?

  3. Goten77
    • 3 years ago
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    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

  4. chihiroasleaf
    • 3 years ago
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    what is erc(x) ?

  5. UnkleRhaukus
    • 3 years ago
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    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:\[\operatorname{erfc}(x)=1-\operatorname {erf}(x)\] ___\[\operatorname {erc}(x)=e^{x^2}\operatorname{erfc(x)}\]

  6. UnkleRhaukus
    • 3 years ago
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    i guess it is the normalised complementary error function

  7. chihiroasleaf
    • 3 years ago
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    ahh I see..... what's your problem? I think you've done it..

  8. UnkleRhaukus
    • 3 years ago
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    yay

  9. UnkleRhaukus
    • 3 years ago
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    so i haven't made any errors?

  10. chihiroasleaf
    • 3 years ago
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    It seems correct..

  11. sirm3d
    • 3 years ago
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    \[\Large {y=\operatorname{erc}(x)=e^{x^2}\left[1-\frac{2}{\sqrt \pi}\int_0^xe^{-u^2}\;\mathrm du\right]\\\frac{dy}{dx}=2xe^{x^2}\left[1-\frac{2}{\sqrt \pi}\int_0^xe^{-u^2}\;\mathrm du\right]+e^{x^2}\left[-\frac{2}{\sqrt \pi}e^{-x^2}\right]\\=2xy-\frac{2}{\sqrt \pi}}\]

  12. sirm3d
    • 3 years ago
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    it's shorter if you apply the fundamental theorem of calculus

  13. UnkleRhaukus
    • 3 years ago
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    i though i did

  14. sirm3d
    • 3 years ago
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    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\[\frac{\mathrm d}{\mathrm dx} \int_a^x f(u)\;\mathrm du=f(x)\]

  15. UnkleRhaukus
    • 3 years ago
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    the only difference i can see your method is that you used erf where i used erfc

  16. UnkleRhaukus
    • 3 years ago
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    I can't say I understand the difference between derivative "under the integral sign" & the fundamental theorem of calculus

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