Question

having some problems in simplifying...

Answer 1

Here is the problems:

Answer 2

\[y = e^{(1/x) - \ln x + c1)}\]

Answer 3

I'm given...

Answer 4

y(-1) = -1

Answer 5

I simplified the equation to:

Answer 6

\[y = Ce^{1/x}/x\]

Answer 7

How should I get the C?

Answer 8

Shouldn't C be just e?

Answer 9

@oldrin.bataku @dan815

Answer 10

What is that Cl or c1 or whatever it is?

Answer 11

The answer in the back of the book is written to be:

Answer 12

\[y= e^{-(1+1/x)}/x \]

Answer 13

1st constant of integration

Answer 14

looks good i think

Answer 15

the book answer?

Answer 16

\[y = e^{(1/x) - \ln x + c1)}\]
\[=e^{\frac{1}{x}}\times e^{-\ln(x)}\times e^{c_1}\]
\[=\frac{e^{\frac{1}{x}}\times e^{c_1}}{e^{\ln(x)}}=\frac{e^{\frac{1}{x}+c_1}}{x}\]

Answer 17

that is what i get anyway. not sure what the book gets or why

Answer 18

I'm so confused. \(y=e^{1/x-\log x+c_1}\) is undefined for \(x=-1\).

Answer 19

well im trying to find C form this equation: \[y = C e^{1/x}/x\] given y(-1)=-1

Answer 20

well that is a good point , isn't it. although \[\frac{e^{\frac{1}{x}+c_1}}{x}\]is not

Answer 21

I am kind of stuck at this point:

Answer 22

\[-1 = C e^{-1}/-1\]

Answer 23

From this I should clearly get "e"

Answer 24

for C

Answer 25

what do you think?

Answer 26

I am thinking book messed up or I copied something wrong!

Answer 27

well thank you all that took the time to reply!

Answer 28

who*

Answer 29

i think you are maybe right, but the book has a C in the exponent

Answer 30

My answer is coming out to be:

Answer 31

\[f(x)=\frac{e^{\frac{1}{x}+c}}{x}\]
\[f(-1)=-1=\frac{e^{-1+c}}{-1}\]
\[e^{-1+c}=1\]
\[c=1\]

Answer 32

\[y = e^{(1+x)/x}/x\]

Answer 33

hmmm i get the same thing
wonder where the minus sign comes from
everything is written correctly, yes?

Answer 34

it isn't for example \(y(1)=-1\) or \(y(-1)=1\)

Answer 35

well I guess I can check if I copied the problem right later. Thanks anyways!

Answer 36

that's what I was thinking.

Answer 37

yw

Answer 38

$$y=e^{1/x-\log x+c_1}=e^{1/x}e^{-\log x}e^{c_1}=\frac{C}xe^{1/x}\\-1=-Ce^{-1}=-\frac{C}e\\C=e$$

Answer 39

yeah I am getting that but the book has the final answer is giving me answer indicating that c should be e^(-1).

Answer 40

ill check later what I copied wrong (if I copied wrong). but thanks anyways!