A community for students.
Here's the question you clicked on:
 0 viewing
UnkleRhaukus
 2 years ago
Show that \(\operatorname{erc}(x)\) satisfies the differential equation
\[\frac{\mathrm dy}{\mathrm dx}=2xy\frac2{\sqrt \pi}\] With \(y=1\) when \(x=0\)
UnkleRhaukus
 2 years ago
Show that \(\operatorname{erc}(x)\) satisfies the differential equation \[\frac{\mathrm dy}{\mathrm dx}=2xy\frac2{\sqrt \pi}\] With \(y=1\) when \(x=0\)

This Question is Closed

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

Goten77
 2 years ago
Best ResponseYou've already chosen the best response.0im 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

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

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0i guess it is the normalised complementary error function

chihiroasleaf
 2 years ago
Best ResponseYou've already chosen the best response.0ahh I see..... what's your problem? I think you've done it..

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0so i haven't made any errors?

chihiroasleaf
 2 years ago
Best ResponseYou've already chosen the best response.0It seems correct..

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.2\[\Large {y=\operatorname{erc}(x)=e^{x^2}\left[1\frac{2}{\sqrt \pi}\int_0^xe^{u^2}\;\mathrm du\right]\\\frac{dy}{dx}=2xe^{x^2}\left[1\frac{2}{\sqrt \pi}\int_0^xe^{u^2}\;\mathrm du\right]+e^{x^2}\left[\frac{2}{\sqrt \pi}e^{x^2}\right]\\=2xy\frac{2}{\sqrt \pi}}\]

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.2it's shorter if you apply the fundamental theorem of calculus

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.2you 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\[\frac{\mathrm d}{\mathrm dx} \int_a^x f(u)\;\mathrm du=f(x)\]

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0the only difference i can see your method is that you used erf where i used erfc

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0I can't say I understand the difference between derivative "under the integral sign" & the fundamental theorem of calculus
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.