Question

if dy/dx = e^y and y = 0 when x = 1, then

Answer 1

if \[\frac{ dy }{ dx }= e ^{y}\]and y = 0 when x = 1, then

Answer 2

a) \[e ^{-y}=x-2\]
b) \[e ^{-y}=2-x\]
c) \[y=\ln|2-x|\]
d) \[y=-\ln|x|\]
e) \[y=\ln[x]\]

Answer 3

First off we need to solve for y as a function of x. All we really know is that e^y is the derivative of y with respect to x. So we need to separate it apart and integrate like this: \[\frac{dy}{dx}=e^y\]separating...\[\int\limits e^{-y}dy=\int\limits dx\] Once you integrate you will get a constant. Solve for this constant by plugging in the point (1,0) that they give you. Then solve for y, and you should have your answer!

Answer 4

wait, that makes no sense, why is the e^-y with dy, and y is it negative,? shouldnt it be \[\int\limits_{}^{}e ^{y}dx = \int\limits_{}^{}dy\]

Answer 5

No, because I divided both sides by e^y to separate the y's from the x's.

Answer 6

oh, i get it, ty