Question

Find the soln. to the initial value problem e^x^2=xe^y with the initial condition y(0)=0

Answer 1

I got the variables separated, but I have no idea how to integrate this...

Answer 2

hmm shouldn't there be a dy/dx somewhere
plugging in (0,0) doesn't work for equation you posted

Answer 3

oh right ! oh im srry, there's
e^x^2 (y')=xe^y

Answer 4

right thats better

Answer 5

:P

Answer 6

ok so you have
\[\int\limits e^{-y} dy = \int\limits x e^{-x^2} dx\]

Answer 7

yeah

Answer 8

left side easy .... integral of e^-y = -e^-y

Answer 9

ohhhh right~ ok haha I hd it as a fraction

Answer 10

right side use substitution

u = -x^2

du = -2x dx --> dx = -du/2x

Answer 11

haha yeah for integrating its usually best to get rid of fractions and use neg exponents

Answer 12

ok so I've got -y on one side now, and ln[(e^-x^2)/2]....how to deal with the right side?

Answer 13

OH WAIT YOU CAN SPLIT IT

Answer 14

oh wait i forgot about the c

Answer 15

how do I deal with the c??? TToTT

Answer 16

Are you still there?

Answer 17

\[e^{-y} = \frac{1}{2} e^{-x^2} +\frac{C}{2}\]
\[-y = \ln \frac{e^{-x^2} +C}{2}\]

Answer 18

*magic* ok so now we just plug in a dnsolve for c?

Answer 19

right

Answer 20

could you take the ln2 out, or would that be unwise and a weird hassle?

Answer 21

yes you could but i would wait til the end, its not needed to find C

Answer 22

....how so?

Answer 23

I feel like it might be easier to deal with the constant from this point:\[\Large\rm e^{-y} = \frac{1}{2} e^{-x^2} +c\]But maybe that's just me.
Make sure you understand how to read function notation:\[\Large\rm y(0)=0 \quad means: \quad y=0 \quad when \quad x=0\]So just plug 0 in for both to solve for c.

Answer 24

x=0, y=0
e^-y = e^0 = 1
e^-x^2 = e^0 = 1
\[1 = \frac{1+C}{2}\]

Answer 25

oh ok so you do need the 2 there~

Answer 26

yeah exactly ^^
i have a bad habit of trying to solve the whole thing first before finding C

Answer 27

>o< haha that doesn't work out well, does it? ^^ thank you~!!! I'm gonna head in now~ you were awesome XD

Answer 28

yw :)