anonymous
  • anonymous
solve dy/dx = e^{x+y} (x+y)^{-1} - 1 y ????
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
solve what
anonymous
  • anonymous
\[\frac{ dy }{ dx } = e^{x+y} (x+y)^{-1} - 1 \]
anonymous
  • anonymous
dame i am not at a level like this one, i am sorry

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

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