anonymous
  • anonymous
having some problems in simplifying...
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Here is the problems:
anonymous
  • anonymous
\[y = e^{(1/x) - \ln x + c1)}\]
anonymous
  • anonymous
I'm given...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
y(-1) = -1
anonymous
  • anonymous
I simplified the equation to:
anonymous
  • anonymous
\[y = Ce^{1/x}/x\]
anonymous
  • anonymous
How should I get the C?
anonymous
  • anonymous
Shouldn't C be just e?
anonymous
  • anonymous
@oldrin.bataku @dan815
Mertsj
  • Mertsj
What is that Cl or c1 or whatever it is?
anonymous
  • anonymous
The answer in the back of the book is written to be:
anonymous
  • anonymous
\[y= e^{-(1+1/x)}/x \]
anonymous
  • anonymous
1st constant of integration
anonymous
  • anonymous
looks good i think
anonymous
  • anonymous
the book answer?
anonymous
  • anonymous
\[y = e^{(1/x) - \ln x + c1)}\] \[=e^{\frac{1}{x}}\times e^{-\ln(x)}\times e^{c_1}\] \[=\frac{e^{\frac{1}{x}}\times e^{c_1}}{e^{\ln(x)}}=\frac{e^{\frac{1}{x}+c_1}}{x}\]
anonymous
  • anonymous
that is what i get anyway. not sure what the book gets or why
anonymous
  • anonymous
I'm so confused. \(y=e^{1/x-\log x+c_1}\) is undefined for \(x=-1\).
anonymous
  • anonymous
well im trying to find C form this equation: \[y = C e^{1/x}/x\] given y(-1)=-1
anonymous
  • anonymous
well that is a good point , isn't it. although \[\frac{e^{\frac{1}{x}+c_1}}{x}\]is not
anonymous
  • anonymous
I am kind of stuck at this point:
anonymous
  • anonymous
\[-1 = C e^{-1}/-1\]
anonymous
  • anonymous
From this I should clearly get "e"
anonymous
  • anonymous
for C
anonymous
  • anonymous
what do you think?
anonymous
  • anonymous
I am thinking book messed up or I copied something wrong!
anonymous
  • anonymous
well thank you all that took the time to reply!
anonymous
  • anonymous
who*
anonymous
  • anonymous
i think you are maybe right, but the book has a C in the exponent
anonymous
  • anonymous
My answer is coming out to be:
anonymous
  • anonymous
\[f(x)=\frac{e^{\frac{1}{x}+c}}{x}\] \[f(-1)=-1=\frac{e^{-1+c}}{-1}\] \[e^{-1+c}=1\] \[c=1\]
anonymous
  • anonymous
\[y = e^{(1+x)/x}/x\]
anonymous
  • anonymous
hmmm i get the same thing wonder where the minus sign comes from everything is written correctly, yes?
anonymous
  • anonymous
it isn't for example \(y(1)=-1\) or \(y(-1)=1\)
anonymous
  • anonymous
well I guess I can check if I copied the problem right later. Thanks anyways!
anonymous
  • anonymous
that's what I was thinking.
anonymous
  • anonymous
yw
anonymous
  • anonymous
$$y=e^{1/x-\log x+c_1}=e^{1/x}e^{-\log x}e^{c_1}=\frac{C}xe^{1/x}\\-1=-Ce^{-1}=-\frac{C}e\\C=e$$
anonymous
  • anonymous
yeah I am getting that but the book has the final answer is giving me answer indicating that c should be e^(-1).
anonymous
  • anonymous
ill check later what I copied wrong (if I copied wrong). but thanks anyways!

Looking for something else?

Not the answer you are looking for? Search for more explanations.