## 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 $$x=0$$

1. UnkleRhaukus


2. UnkleRhaukus

y/n ?

3. Goten77

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

what is erc(x) ?

5. UnkleRhaukus

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

i guess it is the normalised complementary error function

7. chihiroasleaf

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

8. UnkleRhaukus

yay

9. UnkleRhaukus

so i haven't made any errors?

10. chihiroasleaf

It seems correct..

11. sirm3d

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

12. sirm3d

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

13. UnkleRhaukus

i though i did

14. sirm3d

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

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

16. UnkleRhaukus

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

Find more explanations on OpenStudy