Question

HELP! linear algebra & differential equations problem!

Answer 1

i want to know why he did e^-x2 1/(-x^2)' + C

Answer 2

this is for the first order linear equation

Answer 3

oh I see what happened, yo changed the integrating factor for some reason...

Answer 4

no... I was right the first time, you are stuck

Answer 5

\[\int e^{-x^2}dx\]cannot be integrated in terms of simpler functions

Answer 6

the \[\frac1{(x^2)'}=\frac1{2x}\]thing was an attempt to save face :P

Answer 7

it makes no sense and is wrong...

Answer 8

so that part is wrong?

Answer 9

yeah, that DE as no simple solution; it will involve the error function for that integral erf(x)

Answer 10

hm.. how the hell did he even get that then.. 0.o

Answer 11

ya don't...

Answer 12

\[e^{-x^2}y=\int e^{-x^2}dx=\frac12\sqrt{\frac\pi2}\text{erf}(x)+C\]

Answer 13

not really the kind of thing ou cover in introductory diffeQ's

Answer 14

you*

Answer 15

wow... how am i suppose to know that it will become this.. this is insane..

Answer 16

there is most likely a typo lost in translation somewhere
where did you get that [a[er from; do you have the original problem?

Answer 17

paper*

Answer 18

yea its y'-2xy =1, y(a) = b

Answer 19

I had a similar problem in my book but upon closer review it was dx/dy not dy/dx

Answer 20

nvm youd get the same thing lol

Answer 21

\[y(x)=\frac{e^{x^2}}2\left(\sqrt{\frac\pi2}\text{erf}(x)+C\right)\]\[y(a)=\frac{e^{x^2}}2\left(\sqrt{\frac\pi2}\text{erf}(a)+C\right)=b\]

Answer 22

that is the same thing

Answer 23

yea! i see it but i guess he didnt solve it all the way haha =) he didnt know either

Answer 24

\[\int e^{-x^2}dx=\frac12\sqrt{\frac\pi2}\text{erf}(x)+C\]

Answer 25

yeah they don't expect you to know about the error function yet, so that makes sense

Answer 26

thanks so much =) i understand! haha he left it just like integrate e^-x^2 dx

Answer 27

no prob :)\
I would have tried the same I'm sure had I not seen this integral before :P

Answer 28

slight typo\[y(x)=e^{x^2}\left(\sqrt{\frac\pi2}\text{erf}(x)+C\right)\]\[y(a)=e^{x^2}\left(\sqrt{\frac\pi2}\text{erf}(a)+C\right)=b\]added and extra 1/2 :P

Answer 29

do you know how he got \[y=e ^{x^2}(c + \int\limits_{}^{}e ^{-x^2} dx)\] ???

Answer 30

yeah, he just stopped a step earlier....

Answer 31

\[ye^{-x^2}=\int e^{-x^2}dx\implies y=e^{x^2}\int e^{-x^2}dx\]hm... it actually looks like he cheated on the C part lol, it shouldn't be there until after the integration...

Answer 32

...and since\[\int e^{-x^2}dx=\sqrt{\frac\pi2}\text{erf}(x)+C\]I'd say my answer is really more appropriate

Answer 33

oh i see! how about when we plug in y(a) = b how did he get \[y=e ^{x^2} (b ^{-a^2} + \int\limits_{a}^{b} e ^{-t^2} dt \] where did the letter t come from?

Answer 34

hm...

Answer 35

\[y=e^{x^2}\int e^{-x^2}dx\]we really can't introduce C yet technically because we have not integrated, but let's just imagine that that's what they want you to do (I think that is a no-no, but whatever...)

Answer 36

i see dont see how he go a t..... out of nowhere.

Answer 37

oh I though that was your typo lol
okay, now I'm starting to think your teacher is missing a few screws...

Answer 38

i just wanna cry...

Answer 39

\[y(a)=e^{a^2}\left(\int e^{-x^2}dx+C\right)=b\]\[\int e^{-x^2}dx+C=e^{-a^2}b\]yeah I really don't see how we can find C from this....
any ideas silent observer @experimentX ?

Answer 40

you guys are right. find the value of c, by plugging x=a, f(x) = b
http://www.wolframalpha.com/input/?i=y%27+-+2xy+%3D+5

Answer 41

yeah but how do you get his teachers answer?
(not even gonna ask where the t came from lol)

Answer 42

i have no idea how he did this maybe a typo?

Answer 43

the t must be a typo

Answer 44

yea i guess it just possibly cant be explained .. =(

Answer 45

I don't even understand why he can put a and b as the bounds of integration either.... very strange.
I'm gonna have to call it a night, good luck!

Answer 46

thank you so much!!

Answer 47

probably you made some typo check your question again
http://www.wolframalpha.com/input/?i=y%27+-+2xy+%3D+%282x^2%2B1%29%2F%282x^2%29

Answer 48

thanks =)

Answer 49

you are welcome!