Question

this is differential equation & linear algebra level question.

Exact Equation:
Find the general solution, or find a relation that defines the solution implicitly.

ex: (x^2-y)y' + 2x^3 +2xy = 0

Answer 1

first step.. change y' into dy/dx

\[\implies (x^2 - y)\frac {dy}{dx} + 2x^3 + 2xy = 0\]
now multiply dx to all sides
\[\implies (x^2 - y) dy + (2x^3 + 2xy)dx = 0\]
following so far?

Answer 2

yes

Answer 3

now.. this is in teh form Mdx + Ndy = 0 and is exact... agree?

the next step would be to integrate Mdx \[\implies \int Mdx\]
in this case, our M is (2x^3 + 2xy) so our integral becomes \[\implies \int (2x^3 + 2xy)dx\]
got it?

Answer 4

yea i got it up to here.

Answer 5

okay.. now if we simplify the integral...it becomes \[\frac{2x^4}{4} + \frac{2x^2 y}{2} + c\]
so if you simplify that...
\[\implies \frac{x^4}{2} + x^2 y + c\]

Answer 6

now...the next step would be to replace c with g(y)<--g(y) is a function of y

so we have.. \[\implies \frac{x^4}{2} + x^2 y + g(y)\]

Answer 7

im following you! omg! then what happens?

Answer 8

now... you derive that in terms of y

so it becomes \[0 + x^2 + g^\prime(y)\]
still okay?

Answer 9

yes!!!

Answer 10

so now you equate this to N

in our case, N is (x^2 - y) <--because that's the function beside dy

so we have
\[\implies x^2 + g^\prime (y) = x^2 - y\]
you can subtract both sides by x^2

that would give you..
\[g^\prime (y) = -y\]
right?

Answer 11

omg now i see! so what would be the final solution?

Answer 12

you integrate both sides
\[\implies \int g^\prime (y) = \int -y\]
simplify it..
\[g(y) = -\frac{y^2}{2} + c\]

now... remember the earlier expression? \[\frac{x^4}{2} + x^2 y + g(y)\]
you substitute g(y) = -y^2/2 + c here

so you have \[\implies \frac{x^4}{2} + x^2 y - \frac{y^2}{2} + c\]
got it?

Answer 13

so that would be the final solution? because the solution says something like \[y=x^2\pm \sqrt{2x^4+c}\]

Answer 14

hmm seems this answer expressed it in a different way. let me try finding how to get it

Answer 15

thank you so much =)

Answer 16

where did you get that solution?

Answer 17

its actually my teachers solution he made it himself. that is the reason i am so confused? i mean i would say its a typo if its little bit off but it looks like a completely different answer...

Answer 18

okay i got it..

Answer 19

this is the hw =)

Answer 20

\[\frac{x^4}{2} + x^2y - \frac{y^2}{2} + c = 0\]
multiply all terms by 2
\[\implies x^4 + 2x^2y - y^2 + c = 0\]
transpose 2x^2y - y^2
\[\implies x^4 + c = y^2 - 2x^2 y\]
add x^4 to both sides
\[\implies x^4 + c + x^4 = y^2 - 2x^2 y + x^4\]
combine like terms
\[\implies 2x^4 + c = y^2 - 2x^2 y + x^4\]
facotr out..
\[\implies 2x^4 + c = (y - x^2)^2\]
take the square root of both sides
\[\implies \pm \sqrt{2x^4 + c } + y - x^2\]
add x^2 to both sides
\[\implies y = x^2 \pm \sqrt{2x^4 + c}\]
that's some filthy manipulation lol

Answer 21

that second to the last line should be \[\implies \pm \sqrt{2x^4 + c} = y - x^2\]
minor typo

Answer 22

wow thats crazy... why would he do this to me .... =(

Answer 23

calculus teachers are sadistic

Answer 24

anyway...do you get the solution?

Answer 25

basically my teacher wants to solve y= to something so i guess he wanted it ALL the way

Answer 26

but yeah thank you so much you are my life saver without you i would automatically fail this course =)

Answer 27

lol welcome. it's a great review for me too

Answer 28

I guess my way sort of less painful:
y² = x^4 + 2x²y + c
-> y² - 2x²y = x^4 + c
=> ( y - x² ) ² = 2x^4 + c
=> y - x² = ± √ ( 2x^4 + c )
Thus y = x² ± √ ( 2x^4 + c )

Answer 29

wasn't that what i did too?

Answer 30

Might be the detailed explanation makes it looks long :P

Answer 31

oh lol. i'll try to shorten it next time

Answer 32

@lgbasallote No, you're always so cool at explanation. It's great benefit for the asker!

Answer 33

i really like your explanation +)