Question

How to find the x-intersects of this equation?

Answer 1

the problem was to solve this differential equation\[D.E.:xdx+2ye^{-x}dy=0,y(0)=\frac{1}{2}\]After solving, the general solution was \[e^x(x-1)+y^2=c\]then the particular solution\[y=\frac{\sqrt{-4e^x(x-1)-3}}{2}\]In the solution it says it is in the interval \[-0.96<x<0.58\]

How do I find\[-0.96<x<0.58\]?

Answer 2

Here's what it looks like visually.

Answer 3

My attempt at solving for the two x intersects \[y=\frac{\sqrt{-4e^x(x-1)-3}}{2}\]\[y=\frac{\sqrt{4e^x(1-x)-3}}{2}\]\[when\,\,y=0\]\[0=\frac{\sqrt{4e^x(1-x)-3}}{2}\]\[0=4e^x(1-x)-3\]\[\frac{3}{4}=e^x-xe^x\]\[\ln(\frac{3}{4})=\ln(e^x-xe^x)\]
and I get stuck here. The solution on wolfram involves the lambert function but I do not know how to use it.

Answer 4

the approach:
for 'y' to be real,
the quantity under square root must be positive.

\(-4e^x(x-1)-3 >0\)

now this gives what you wanted.
https://www.wolframalpha.com/input/?i=-4e^x+%28x-1%29+-3%3E0

bot really sure how to get it by hand

Answer 5

*but not really sure ..

Answer 6

Yeah. I'm planning on skipping this. Since it's too advanced for me, still have a lot to do and I wasted an hour on this xD

Answer 7

right, keep this question open, so that if any1 has an approach, they can help.

and move on :)

Answer 8

Thanks for trying though. I'll probably leave this for a while longer till I close this thread.

Answer 9

i was thinking, can we use linear approximation for e^x
and get a quadratic on left...to get its 2 roots as -0.96, and 0.58

Answer 10

Ah, is that the one where we approximate by using differentials? I forgot how to use that. It's still in my notebook though. Though I have no idea sorry. Also I'll be afk hartnn, I appreciate the thought.

Answer 11

\(e^x \approx 1+x\)

Answer 12

(-0.5, 0.5)
doesn't work :P

Answer 13

we'd get closer to actual answer,
if we linearly approximate
e^x(x-1) = (1-x^2/2)

-4(-1+x^2/2) -3 >0 gives -0.707 < x < 0.707