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gerryliyanaBest ResponseYou've already chosen the best response.0
\[\frac{ dy }{ dx } = e^{x+y} (x+y)^{1}  1 \]
 one year ago

MikiaseKebedeBest ResponseYou've already chosen the best response.0
dame i am not at a level like this one, i am sorry
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
ok..., no problem @MikiaseKebede nevermind :)
 one year ago

MikiaseKebedeBest ResponseYou've already chosen the best response.0
you should ask http://openstudy.com/users/jim_thompson5910
 one year ago

nubeerBest ResponseYou've already chosen the best response.0
i think you have to apply the product rule
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
what did u get it??
 one year ago

nubeerBest ResponseYou've already chosen the best response.0
hmm i havent solved it just trying to help you with it
 one year ago

nubeerBest ResponseYou've already chosen the best response.0
or there is another easier way u =x+y du/dx = 1 du =dx
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
http://mathhelpforum.com/differentialequations/194955solvingdydxexy.html
 one year ago

mayankdevnaniBest ResponseYou've already chosen the best response.0
solve dy/dx = e^{x+y}
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
@mayankdevnani No, i mean \[\frac{ dy }{ dx } = e^{x+y} (x+y)^{1}  1 \]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
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 :)
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
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?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
emm...there is a little mistake in ur separation\[ze^{z}\text{d}z=\text{d}x\]
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
\[\int\limits \frac{ z }{ e^{z} } dz = \int\limits dx \]
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
\[\frac{ 1 + z }{ e^{z} } = x\] \[\frac{ 1 + x + y }{ e^{x+y} } = x\] \[\frac{ x+y1 }{ x } = e^{x+y}\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
integration by parts for z :)
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
\[\int\limits ze^{z} dz = \int\limits dx\] 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 ??
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
ln e = 1 re substitute z as x+y
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
\[ze^{z}  e^{z} = x\] \[(x+y)e^{(x+y)}  e^{(x+y)} = x\] \[(xy  1) e^{(x+y)} = x\] \[(1+x+y) = x e^{(x+y)}\]
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
did i miss something?
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
nopes, thats correct. :)
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
i don't think, y can be isolated....
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=%281%2Bx%2By%29%3D%E2%88%92xe%5E%28x%2By%29+solve+for+y
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
what it means to "W is the product log function"??
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
even idk that... not a standard function for sure....
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
ok Thank you all so much! i really appreciate it :)
 one year ago
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