Growth and Decay

**holly123** · January 8th 2009, 12:02 PM

1.) xy dx + e^(-x^2)dy=0 ; y(o)=1
i got that problem down to 1/2 e^(-x^2) = 1/2(y^2) - lny
but i don't know how to solve that for y

thanks!!

**Soroban** · January 8th 2009, 02:43 PM

Hello, holly123!

You've made an error somewhere . . .

$1)\;\;xy\,dx + e^{-x^2}dy\:=\:0,\quad y(0)\:=\:1$

Separate variables:
. . $e^{-x^2}dy \:=\:-xy\,dx \quad\Rightarrow\quad \frac{dy}{y} \:=\:-xe^{x^2}\,dx$

Integrate: . $\ln y \:=\:-\tfrac{1}{2}e^{x^2} + C$ .[1]

. . Since $y(0) = 1\!:\;\; \ln1 \:=\:-\tfrac{1}{2}e^0 + C \quad\Rightarrow\quad C \:=\:\tfrac{1}{2}$

Then [1] becomes: . $\ln y \:=\:-\tfrac{1}{2}e^{x^2} + \tfrac{1}{2} \quad\Rightarrow\quad \ln y \:=\:\tfrac{1}{2}\left(1 - e^{x^2}\right)$

Therefore: . $y \;=\;e^{\frac{1}{2}(1-e^{x2})}$

**holly123** · January 8th 2009, 07:18 PM

oh im sorry i wrote it wrong! it's
xy dx + e^(-x^2) (y^2-1) dy=0

sorry im confused on how to use the symbols and write this all out correctly and i completely forgot to add the (y^2 -1)

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