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gerryliyana
 3 years ago
solve dy/dx = e^{x+y} (x+y)^{1}  1
y ????
gerryliyana
 3 years ago
solve dy/dx = e^{x+y} (x+y)^{1}  1 y ????

This Question is Closed

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

MikiaseKebede
 3 years ago
Best ResponseYou've already chosen the best response.0dame i am not at a level like this one, i am sorry

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.0ok..., no problem @MikiaseKebede nevermind :)

MikiaseKebede
 3 years ago
Best ResponseYou've already chosen the best response.0you should ask http://openstudy.com/users/jim_thompson5910

nubeer
 3 years ago
Best ResponseYou've already chosen the best response.0i think you have to apply the product rule

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.0what did u get it??

nubeer
 3 years ago
Best ResponseYou've already chosen the best response.0hmm i havent solved it just trying to help you with it

nubeer
 3 years ago
Best ResponseYou've already chosen the best response.0or there is another easier way u =x+y du/dx = 1 du =dx

mayankdevnani
 3 years ago
Best ResponseYou've already chosen the best response.0http://mathhelpforum.com/differentialequations/194955solvingdydxexy.html

mayankdevnani
 3 years ago
Best ResponseYou've already chosen the best response.0solve dy/dx = e^{x+y}

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

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.4let \[z=x+y\]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 :)

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.0ok @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?

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

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \frac{ z }{ e^{z} } dz = \int\limits dx \]

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 + z }{ e^{z} } = x\] \[\frac{ 1 + x + y }{ e^{x+y} } = x\] \[\frac{ x+y1 }{ x } = e^{x+y}\]

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.4integration by parts for z :)

gerryliyana
 3 years ago
Best 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 ??

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0ln e = 1 re substitute z as x+y

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

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.0did i miss something?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0nopes, thats correct. :)

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0i don't think, y can be isolated....

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

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.0what it means to "W is the product log function"??

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0even idk that... not a standard function for sure....

gerryliyana
 3 years ago
Best ResponseYou've already chosen the best response.0ok Thank you all so much! i really appreciate it :)
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