Question

help me in differential equation topic is exact. will give medal and fan

Answer 1

question: use method III to solve this problem in exact in DE \[ye ^{xy} dx + xe ^{xy} dy = 0\]

Answer 2

final answer should be \[xy = C\]

Answer 3

check whether it is exact

Answer 4

\[\frac{ dm }{ dy } = d(ye ^{xy}) \frac{ dn }{ dx } = d(xe ^{xy})\]

Answer 5

so what are their derivatives?
are they equal?

Answer 6

y = EXACT

Answer 7

formula for METHOD 3 IS \[\int\limits_{0}^{X} M(t,y) dt + \int\limits_{0}^{y} N (0,t) dt = c\]

Answer 8

\[\int\limits_{y=constant} Mdx+\int\limits (Terms~of~N~free~from~x~term)dy=C\]

Answer 9

are my derivatives right/

Answer 10

@rvc

Answer 11

oh m back

Answer 12

u need to integrate with respect to x

Answer 13

can u do the method 3 and i do the method 5

Answer 14

for the M term : keeping y as constant

Answer 15

method 3
may i know which method you are exactly referring to?

Answer 16

methd 3 formula is \[\int\limits_{0}^{x} M (t,y)dt + \int\limits_{0}^{y} N (0,t) dt = C\]

Answer 17

well you just need to integrate the M term

Answer 18

and method 5 is \[\int\limits M (x,y)dx + \int\limits N(x,y) dy = c\]

Answer 19

\[\int\limits y\cdot~e^{xy}~dx\]

Answer 20

\[\frac{ y^2 }{ 2 }\]

Answer 21

i am not familiar with method 3
im familiar with method 5

Answer 22

oh, i give u sample

Answer 23

remember : y is a constant

Answer 24

as per method 5

Answer 25

@Directrix @phi may help u
i can guide u with the method 5

Answer 26

here

Answer 27

u see it?

Answer 28

lemme see

Answer 29

i sent a file

Answer 30

i watched

Answer 31

so here basically in the M term
x is replaced by t and y is kept as it is

Answer 32

lemme see ur work, then i do the method 4 and 5

Answer 33

the ans by method 5 is : ?

Answer 34

xy = c

Answer 35

i guess its e^{xy}=C

Answer 36

correct
because thats what i got

Answer 37

but when un ln^ e(xy) its xy

Answer 38

can i see