Question

this is differential equation & linear algebra question:
im trying to looking for exact equation
(x^2-y)y'+2x^3+2xy=0
the answer needs to be 3.) y=x^2 +_ (2x^4+c)^1/2

Answer 1

integrate one side up, add a generic constant function to it; and derivative it down wrt the other variable. Then compare to find the derivative of the generic function

Answer 2

can you steps? i cant really see it with just writing

Answer 3

\[(x^2-y)dy+(2x^3+2xy)dx=0\]

\[\int (x^2-y)dy=x^2y-\frac12y^2+g(x)\]
\[\frac{d}{dx}(x^2y-\frac12y^2+g(x))=2xy+g'(x)\]
\[\cancel{2xy}+g'(x)=2x^3+\cancel{2xy}\]

Answer 4

integrate g' to determine g and fill it in; then check the work :)

Answer 5

here what i got i did My = 2x^3 + 2xy then did partial derivate respect to y then got 2x then Nx = (x^2-y) and got 2x also which is same so we can conclude that it is exact equation and then i did Mdx = 2x^3 +2xy which gives 1/2x^4+x^2y + g'(y)

Answer 6

why did you do M wrt y?

Answer 7

M(x,y) =
N(x,y) =

Answer 8

if we apply your workings
\[(x^2-y)y'+2x^3+2xy=0\]
\[(x^2-y)\frac{dy}{dx}+2x^3+2xy=0\]
\[(x^2-y)dy+(2x^3+2xy)dx=0\]
\[Ndy+Mdx=0\]
M is not integrated wrt y

Answer 9

i read you wrong :) you were proving it was exact

Answer 10

yeah

Answer 11

and then i did integrate to Mx

Answer 12

it doesnt matter which side you start integrating; i started on the "N" in this case

Answer 13

okay but i dont seem to get the answer provided by my teacher

Answer 14

your setup is going along fine; except, its not g'(y); its just a generic function of y: g(y)

Answer 15

oh okay so then after i did Mx i now integrate it to respect to Ndy then im stuck...

Answer 16

it tends to be easier if you take the derivative down to Ndy to compare it with

Answer 17

instead of using Mdx first?

Answer 18

you already did the "first"