Question

Determine which of the following expressions represent the exact differential and find the corresponding function satisfying the condition

f(2,-1) = 2 - 5ln2

if such function exists.

a) (sin^2(xy/2) + e^x/y) dx - (cos(2x/y) - e^x/y) dy;

b) (1/y^2-5/x) dx + 2/y(5-(x/y^2) dy

Answer 1

an exact diff. is one where you \(\int M(x,y) dx\) and add on a function in y, then \(\frac{d}{dx}\) down into N(x,y)

i think

Answer 2

or vice versa

Answer 3

\[(sin^2(\frac{xy}{2}) + e^{\frac{x}{y}}) dx - (cos(\frac{2x}{y}) - e^{\frac{x}{y}}) dy\]
\[\int sin^2(\frac{xy}{2}) + e^{\frac{x}{y}} dx\]
\[\int sin^2(\frac{xy}{2})dx + ye^{\frac{x}{y}} \]
and what is the reduction formula for sins?

Answer 4

sin^2(x) = 1/2 -cos(2x)/2 i think

Answer 5

\[\int (\frac{1}{2}-\frac{1}{2}cos(2\frac{xy}{2}))dx + ye^{\frac{x}{y}}\]
\[\frac{1}{2}x-\frac{1}{2}y\ sin(xy) + ye^{\frac{x}{y}}\]

Answer 6

err... 1/2y on the sin

Answer 7

\[\frac{d}{dy}\ \frac{1}{2}x-\frac{1}{2y}\ sin(xy) + ye^{\frac{x}{y}}+g(y)\]

\[\frac{d}{dy}(-\frac{1}{2y}\ sin(xy) + ye^{\frac{x}{y}})+g'(y)\]

Answer 8

\[\frac{d}{dy}(-\frac{1}{2y}\ sin(xy))=-\frac{1}{2y}'\ sin(xy)+\frac{-1}{2y}\ sin(xy)' \]
\[=>\ \frac{1}{2y^2}\ sin(xy)-\frac{1}{2y}\ x\ cos(xy) \]

\[\frac{d}{dy}ye^{\frac{x}{y}}=y'e^{\frac{x}{y}}+ye'^{\frac{x}{y}}\]
\[=>\ e^{\frac{x}{y}}+y \frac{x}{y}'e^{\frac{x}{y}}\]
\[=>\ e^{\frac{x}{y}}+y \frac{-x}{y^2}e^{\frac{x}{y}}\]
\[=>\ e^{\frac{x}{y}}-\frac{x}{y}e^{\frac{x}{y}}\]

\[\frac{1}{2y^2}\ sin(xy)-\frac{1}{2y}\ x\ cos(xy) + e^{\frac{x}{y}}-\frac{x}{y}e^{\frac{x}{y}}+g'(y)\]
maybe?

Answer 9

and equate that to the dy side

Answer 10

\[\small\frac{1}{2y^2}\ sin(xy)-\frac{1}{2y}\ x\ cos(xy) + e^{\frac{x}{y}}-\frac{x}{y}e^{\frac{x}{y}}+g'(y)=cos(\frac{2x}{y}) - e^{\frac{x}{y}}\]
what a mess ....

Answer 11

\[\small\frac{sin(xy)-xy\ cos(xy)-2xy\ e^{\frac{x}{y}}+2y^2 e^{\frac{x}{y}}}{2y^2}+g'(y)=\frac{2y^2cos(\frac{2x}{y}) - 2y^2e^{\frac{x}{y}}}{2y^2}\]
\[\small{sin(xy)-xy\ cos(xy)-2xy\ e^{\frac{x}{y}}+2y^2 e^{\frac{x}{y}}}+g'(y)={2y^2cos(\frac{2x}{y}) - 2y^2e^{\frac{x}{y}}}\]
\[\small g'(y)={-sin(xy)+xycos(xy)+2xye^{\frac{x}{y}}-2y^2 e^{\frac{x}{y}}}+{2y^2cos(\frac{2x}{y})-2y^2e^{\frac{x}{y}}}\]
then we either simplify and up that mess or we up it and let the undergrads sort thru it

Answer 12

Thanks for your help would you know what I need to do to complete the question correctly? This is for an assignment for university? I am stuck.

Answer 13

i would have to rework it to make sure i didnt mess any of it up

Answer 14

ok only if you have time. would be appreciated final maths course and i want to pass it.

Answer 15

so far its correct down to the "maybe" part :)

\[\frac{-2 y e^{\frac{x}{y}} (x-y)+sin(x y)-xy\ cos(x y)}{2 y^2}+g'(y)\]

the missing part; the g'(y) is what should make this equal to what is given already in the "dy" part of the problem.

Answer 16

\["that"+g'(y)=cos(2x/y) - e^{x/y}\]
\[g'(y)=cos(\frac{2x}{y}) - e^{x/y}-"that"\]
\[\small g'(y)=cos(\frac{2x}{y}) - e^{x/y}-\left(\frac{-2 y e^{\frac{x}{y}} (x-y)+sin(x y)-xy\ cos(x y)}{2 y^2}\right)\]
\[ g'(y)=cos(\frac{2x}{y}) - e^{x/y}-\frac{2 y e^{\frac{x}{y}} (x-y)-sin(x y)+xy\ cos(x y)}{2 y^2}\]

Answer 17

i spose it would be easier to split it up into parts and work its integration

Answer 18

unless there is an easier way that i cant recall, this thing turns out ugly

Answer 19

we could try to up the dy and bring it down to the dx and see if that is a nicer fit

Answer 20

No, it's easier than all this.

A differential
P dx + Q dy

is exact if thereis a function F such that

dF = P dx + Q dy

I.e.,
\[P = \partial F / \partial x \ \ \text{ and } \ \ Q = \partial F/\partial y\]

So for the first example,

(sin^2(xy/2) + e^x/y) dx - (cos(2x/y) - e^x/y) dy

it is clear that there is no such F because just looking at hte e^x/y terms, there is no function G such that that partial deriv wrt x gives e^x/y and the partial deriv wrt y gives -e^x/y

Formalize that by using the two pdes above if you like.

Now for the second example, isn't written correctly. Can you give it to us again?

Answer 21

yeah, those y denominators in the dy part was causing me concern ....

when i tried to correlate it with the wolf, it spit back it "ugly" answers ...

Answer 22

(1/y^2-5/x) dx + 2/y(5-x/y^2)dy