Question

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

Answer 1

\[
\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*}\]\[
\begin{align*}
y(x)&=\erc x\\
&=e^{x^2}\erfc x\\
&=\ercint x\\
\\
y(0)&=\erfcint 0\\
&=1\\
\\
\de yx&=\frac2{\sqrt \pi}\left[\de{\left(e^{x^2}\right)}x\intl x\infty{e^{-u^2}}u+e^{x^2}\intl x\infty {\pa {\left(e^{-u^2}\right)}u}u\right]\\
&=\frac2{\sqrt \pi}\left[2xe^{x^2}\intl x\infty{e^{-u^2}}u+e^{x^2}\left.e^{-u^2}\right|_x^\infty\right]\\
&=\frac2{\sqrt \pi}\left[2xe^{x^2}\intl x\infty{e^{-u^2}}u-e^{x^2}e^{-x^2}\right]\\
&=2xe^{x^2}\frac{2}{\sqrt \pi}\intl x\infty{e^{-u^2}}u-\frac2{\sqrt \pi}\\
&=2xy-\frac2{\sqrt \pi}\\
\end{align*}
\]

Answer 2

y/n ?

Answer 3

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

Answer 4

what is erc(x) ?

Answer 5

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:\[\operatorname{erfc}(x)=1-\operatorname {erf}(x)\]
___\[\operatorname {erc}(x)=e^{x^2}\operatorname{erfc(x)}\]

Answer 6

i guess it is the normalised complementary error function

Answer 7

ahh I see.....

what's your problem? I think you've done it..

Answer 8

yay

Answer 9

so i haven't made any errors?

Answer 10

It seems correct..

Answer 11

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

Answer 12

it's shorter if you apply the fundamental theorem of calculus

Answer 13

i though i did

Answer 14

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\[\frac{\mathrm d}{\mathrm dx} \int_a^x f(u)\;\mathrm du=f(x)\]

Answer 15

the only difference i can see your method is that you used erf where i used erfc

Answer 16

I can't say I understand the difference between
derivative "under the integral sign" & the fundamental theorem of calculus