Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

UnkleRhaukus

  • 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 \(x=0\)

  • This Question is Closed
  1. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    y/n ?

  3. Goten77
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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

  4. chihiroasleaf
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what is erc(x) ?

  5. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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:\[\operatorname{erfc}(x)=1-\operatorname {erf}(x)\] ___\[\operatorname {erc}(x)=e^{x^2}\operatorname{erfc(x)}\]

  6. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i guess it is the normalised complementary error function

  7. chihiroasleaf
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  8. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yay

  9. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so i haven't made any errors?

  10. chihiroasleaf
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It seems correct..

  11. sirm3d
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  12. sirm3d
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  13. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i though i did

  14. sirm3d
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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\[\frac{\mathrm d}{\mathrm dx} \int_a^x f(u)\;\mathrm du=f(x)\]

  15. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  16. UnkleRhaukus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  17. Not the answer you are looking for?
    Search for more explanations.

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find 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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.