Question

solve dy/dx = e^{x+y} (x+y)^{-1} - 1

y ????

Answer 1

solve what

Answer 2

\[\frac{ dy }{ dx } = e^{x+y} (x+y)^{-1} - 1 \]

Answer 3

dame i am not at a level like this one, i am sorry

Answer 4

ok..., no problem @MikiaseKebede nevermind :)

Answer 5

you should ask http://openstudy.com/users/jim_thompson5910

Answer 6

him

Answer 7

ok

Answer 8

he is good

Answer 9

i think you have to apply the product rule

Answer 10

what did u get it??

Answer 11

hmm i havent solved it just trying to help you with it

Answer 12

or there is another easier way
u =x+y
du/dx = 1
du =dx

Answer 13

http://mathhelpforum.com/differential-equations/194955-solving-dy-dx-e-x-y.html

Answer 14

solve dy/dx = e^{x+y}

Answer 15

@mayankdevnani No, i mean
\[\frac{ dy }{ dx } = e^{x+y} (x+y)^{-1} - 1 \]

Answer 16

let \[z=x+y\]and u have\[\frac{\text{d}z}{\text{d}x}=1+\frac{\text{d}y}{\text{d}x}\]put this in the original equation\[\frac{\text{d}z}{\text{d}x}-1=\frac{e^z}{z}-1\]\[\frac{\text{d}z}{\text{d}x}=\frac{e^z}{z}\]u have a separable differential equation :)

Answer 17

ok @mukushla, then i got
\[\int\limits z dz = \int\limits e^{z} dx\]
\[\frac{ z^{2} }{ 2 } = e^{z} x\]
\[\frac{ (x+y)^{2} }{ 2 } = e^{x+y} x\]
Did I miss something?

Answer 18

emm...there is a little mistake in ur separation\[ze^{-z}\text{d}z=\text{d}x\]

Answer 19

\[\int\limits \frac{ z }{ e^{z} } dz = \int\limits dx \]

Answer 20

exactly

Answer 21

\[\frac{ -1 + z }{ e^{z} } = x\]
\[\frac{ -1 + x + y }{ e^{x+y} } = x\]
\[\frac{ x+y-1 }{ x } = e^{x+y}\]

Answer 22

integration by parts for z :)

Answer 23

\[\int\limits ze^{-z} dz = \int\limits dx\]
u = z, \(dv = e^{-z} dz\), then du = dz, and \(v = -e^{-z}\)
So
\[-ze^{-z} - \int\limits -e^{-z} dz = \int\limits dx\]
\[-ze^{-z} - \frac{ e^{-z} }{ \ln e } = x\]

then ??

Answer 24

ln e = 1
re substitute z as x+y

Answer 25

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

Answer 26

did i miss something?

Answer 27

nopes, thats correct. :)

Answer 28

so how to find y?

Answer 29

i don't think, y can be isolated....

Answer 30

http://www.wolframalpha.com/input/?i=%281%2Bx%2By%29%3D%E2%88%92xe%5E%28x%2By%29+solve+for+y

Answer 31

what it means to "W is the product log function"??

Answer 32

even idk that... not a standard function for sure....

Answer 33

ok Thank you all so much! i really appreciate it :)

Answer 34

welcome ^_^