Question

guys i need help on this exact DE equation:
(x+2y)dx+(2x+3y+1)dy=0

Answer 1

@Hero

Answer 2

so check it first
\[M = x+ 2y => M_y =2\]
\[N=2x+3y+1 => N_y=2\]
Therefore, it's exact.

Answer 3

can i know what is M and what is N?

Answer 4

the answer that i got from my group is kinda different

Answer 5

well your DE is \[(x+2y)dx +(2x+3y+1)d0\]

Answer 6

I haven't gotten to the answer yet lol!

Answer 7

xD ok
sry

Answer 8

\[f(x,y) = \int (x+2y)dx = \frac{x^2}{2}+2xy +h(y)\]
then differentiate \(f(x,y)\) for \(h(y)\) (with respect to \(y\))
\[=> \frac{x^2}{2}+2xy+h(y) => 2x + g'(y)\]
\[=> 2x+ g'(y) = 2x+ 3y +1\]
so \[g'(y) = 3y+1\]
integrating \[g'(y) = 3y+1\] with respect to \(y\) leads to \[g(y) = \int (3y+1)dy = \frac{3}{2}y^2+y\]
so substituting \[\frac{x^2}{2}+2xy+ \frac{3}{2}y^2+y =c_1\]

Answer 9

nice, thnx mimi for answering it, its exactly as the default answer and its logical

Answer 10

pardon? default answer?

Answer 11

i mean the answer which my group got, bah pardon my english, its not my main language

Answer 12

You're actually pretty understandable :D

Answer 13

mimi, are you a lecturer of some sort?

Answer 14

No, she's just awesome :D

Answer 15

:) glad there are people like her

Answer 16

Welcome to Openstudy, @5alood
Stick around, maybe there's somebody with a question you can help with ^.^

Answer 17

thank you, this is really helpful, and i will do my best to help :)