Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Differential Equations
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
solve what
\[\frac{ dy }{ dx } = e^{x+y} (x+y)^{-1} - 1 \]
dame i am not at a level like this one, i am sorry

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

ok..., no problem @MikiaseKebede nevermind :)
him
ok
he is good
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\[ze^{-z}\text{d}z=\text{d}x\]
\[\int\limits \frac{ z }{ e^{z} } dz = \int\limits dx \]
exactly
\[\frac{ -1 + z }{ e^{z} } = x\] \[\frac{ -1 + x + y }{ e^{x+y} } = x\] \[\frac{ x+y-1 }{ x } = e^{x+y}\]
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. :)
so how to find y?
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 :)
welcome ^_^

Not the answer you are looking for?

Search for more explanations.

Ask your own question