Question

Solve the differential equation.
xy' - (x + 1)y = 0

Answer 1

x(dy/dx) - (x+1)y = 0
x(dy/dx) = (x+1)y
dy/dx = [ (x+1)/x ] y

1/y (dy) = (x+1)/x dx ::: integrate

can yu go from there?

Answer 2

somehow i got (x-1)/x

Answer 3

as a final answeR?

Answer 4

noo instead of (x+1)/x dx

Answer 5

my final answer was y = -1/2Ce^x(x)^-2

Answer 6

i have a feeling that's not right

Answer 7

hmm not what i got, wait i'm gonna type what i have

Answer 8

ok now i'm up to ln(y) = x + ln(x) + C.. not sure what to do next

Answer 9

\[xy' - (x + 1)y = 0\]\[x{dy \over dx} - (x+1)y = 0\]\[x{dy \over dx} = (x+1)y\]\[{dy \over dx} = {(x+1) \over x} (y)\]\[{1 \over y} dy = {(x+1) \over x} dx\]
Integrate:\[\ln|y|= \ln|x|+x +C\]\[e^{lny} = e^{lnx+x+C}=e^{lnx}*e^{x}*e^{C}\]\[y = A*x*e^{x}\]
\[y = Axe^{x}\]

that's what i have

Answer 10

is that right?

Answer 11

I think soo.. what is A?

Answer 12

e^C = constant

a constant which i called A, you can called it whatever..

Answer 13

but did yu understand everything?

Answer 14

ohh gotcha.. i understand it.. thank you.. can you help me with other problems too?

Answer 15

good. i sure can, just post them and i'll try

Answer 16

\[\int\limits_{1/2}^{2} \] [(e^1/x) / (x^2)] dx

Answer 17

\[\int\limits_{2}^{2}{{e^{1} \over x} \over x^{2}} dx\]
is that what yu tried to write

Answer 18

\[{e^{1/x} \over x^{2}}\]

Answer 19

that didn't come out on my computer - maybe bc i'm using a mac. but its the integration bounded from 1/2 to 2

Answer 20

and the function that you just wrote

Answer 21

ok

Answer 22

do a product rule between e^(1/x) and x^(-2)

Answer 23

can you do that ?

Answer 24

i'm not really sure.. the integration of e^(1/x) is just e^(1/x)?

Answer 25

-e^(1/x)*x^-1??

Answer 26

the final answer comes up to be e^2 - sqrt(e) i think

Answer 27

how did you get that?