Question

Determine whether the given relation is an implicit solution to the given differential equation: e^xy + y = x - 1 for dy/dx = (e^-xy - y) / (e^-xy + x).
Show all the steps please and why. This is not a test or something like that.

Answer 1

hi nightie

Answer 2

do you wanna see how this be done?

Answer 3

Yes

Answer 4

e^xy + y = x - 1 taking the derivative on both sides we get
e^xy (xdy/dx+ y) +dy/dx= 1
(1/e^-xy)(xdy/dx+ y) = 1-dy/dx
(xdy/dx+ y) = (e^-x)- (e^-x) dy/dx
(xdy/dx)- (e^-x) dy/dx = (e^-x)- y
(dy/dx)(e^-x +x) = (e^-x)- y
dy/dx = [(e^-x)- y]/[(e^-x +x) ] ans

Answer 5

if you have any question dont hesitate to ask me....lol

Answer 6

You have to take the derivative of both sides of e^(xy) + y = x - 1

Answer 7

\[\frac{d}{dx} (e^{xy} + y) = \frac{d}{dx} ( x - 1)\]

Answer 8

For e^(xy) you will use the chain rule because the exponent of e is a product of two functions x and y.

Answer 9

\[e^{xy}(x \frac{dy}{dx} + y) + \frac{dy}{dx} = 1\]

Answer 10

Because the solution you are looking has e^(-xy) divide both sides of equation by e^(xy)

Answer 11

\[x \frac{dy}{dx} + y + e^{-xy} \frac{dy}{dx} = e^{-xy}\]

Answer 12

Now you can collect your like terms keeping in mind that you are solving for dy/dx so you can factor it out on the left hand side:

Answer 13

\[(x + e^{-xy}) \frac{dy}{dx} = e^{-xy} - y\]

Answer 14

Now dividing both sides by (x + e^(-xy)):

Answer 15

\[\frac{dy}{dx}= \frac{e^{-xy}- y}{e^{-xy} + x}\]

Answer 16

Indeed the relation is an implicit solution to the given differential equation.

Answer 17

Nice, thanks!

Answer 18

you're welcome