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gerryliyana
solve dy/dx = e^{x+y} (x+y)^{-1} - 1 y ????
\[\frac{ dy }{ dx } = e^{x+y} (x+y)^{-1} - 1 \]
dame i am not at a level like this one, i am sorry
ok..., no problem @MikiaseKebede nevermind :)
you should ask http://openstudy.com/users/jim_thompson5910
i think you have to apply the product rule
what did u get it??
hmm i havent solved it just trying to help you with it
or there is another easier way u =x+y du/dx = 1 du =dx
http://mathhelpforum.com/differential-equations/194955-solving-dy-dx-e-x-y.html
solve dy/dx = e^{x+y}
@mayankdevnani No, i mean \[\frac{ dy }{ dx } = e^{x+y} (x+y)^{-1} - 1 \]
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 :)
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?
emm...there is a little mistake in ur separation\[ze^{-z}\text{d}z=\text{d}x\]
\[\int\limits \frac{ z }{ e^{z} } dz = \int\limits dx \]
\[\frac{ -1 + z }{ e^{z} } = x\] \[\frac{ -1 + x + y }{ e^{x+y} } = x\] \[\frac{ x+y-1 }{ x } = e^{x+y}\]
integration by parts for z :)
\[\int\limits ze^{-z} dz = \int\limits dx\] 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 ??
ln e = 1 re substitute z as x+y
\[-ze^{-z} - e^{-z} = x\] \[-(x+y)e^{-(x+y)} - e^{(x+y)} = x\] \[(-x-y - 1) e^{-(x+y)} = x\] \[(1+x+y) =- x e^{(x+y)}\]
did i miss something?
nopes, thats correct. :)
i don't think, y can be isolated....
http://www.wolframalpha.com/input/?i=%281%2Bx%2By%29%3D%E2%88%92xe%5E%28x%2By%29+solve+for+y
what it means to "W is the product log function"??
even idk that... not a standard function for sure....
ok Thank you all so much! i really appreciate it :)