Question

find the particular solution satisfying the initial condition indicated.

y' = x e^(y-x^2) when x=0, y=0

Answer 1

I tried using ln, it seems to go nowhere

Answer 2

separation of variables

Answer 3

\[\large y' = x e^{y-x^2} \]
\[\large \dfrac{dy}{dx} = \dfrac{x e^{y}}{e^{x^2}} \]

\[\large \dfrac{1}{e^y}dy = \dfrac{x }{e^{x^2}} dx \]

Answer 4

integrate ^^

Answer 5

wow! I didn't thought of doing that.
thanks!
:)

Answer 6

yw :)

Answer 7

@ganeshie8 just a clarification, is the answer 2e^-y + e^(x^2) = 3 ?

Answer 8

\[\large \dfrac{1}{e^y}dy = \dfrac{x }{e^{x^2}} dx\]

\[\large \int e^{-y}dy = \int x e^{-x^2} dx\]

Answer 9

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

\[\large e^{-x^2}-2e^{-y}= C\]

Answer 10

thats the general solution, right ?

Answer 11

yes

Answer 12

plugin (0, 0)
you get C = 1 - 2 = -1

Answer 13

then the particular solution would be

\[\large e^{-x^2}-2e^{-y}= -1\]

Answer 14

oh, I got the sign incorrectly.

can you change the equation into

y=ln [2/(1+e^-(x^2))]
?
because that's the answer on my book

Answer 15

yes just isolate the y

Answer 16

\[\large e^{-x^2}-2e^{-y}= -1 \]

\[\large e^{-x^2}+1 = 2e^{-y} \]

Answer 17

\[\large e^{-x^2}-2e^{-y}= -1 \]

\[\large \dfrac{e^{-x^2}+1}{2} = e^{-y} \]

\[\large \dfrac{2}{e^{-x^2}+1} = e^{y} \]

Answer 18

take ln both sides

Answer 19

\[\large \ln\left( \dfrac{2}{e^{-x^2}+1}\right) = \ln e^{y} \]

Answer 20

\[\large \ln\left( \dfrac{2}{e^{-x^2}+1}\right) = y\ln e \]

Answer 21

\[\large \ln\left( \dfrac{2}{e^{-x^2}+1}\right) = y \]

Answer 22

ok, Thanks! I understood it. :)