Question

(2x+3)+(2y-2)y'=0
differential problem

Answer 1

we have
\[2x+3+(2y-2)\frac{dy}{dx}=0\]we can write this as
\[(2x+3)dx=-(2y-2)dy\]
let's integrate both sides
\[x^2+3x+C= -(y^2-2y)\]
so the differential equation's solution is
\[y^2-2y+x^2+3x+C=0\]
C is constant of integration

Answer 2

Did you understand Gia?

Answer 3

yes, thanks

Answer 4

welcome:)

Answer 5

how about this
(2xy^2+2y) + (2x^2 y + 2x) y' = 0

Answer 6

Gia I'm gonna solve this but next time use one post for only one question

Answer 7

okay, i'm newbee in here, sorry if i did a mistake

Answer 8

No, problem Gia :D
Welcome to Open Study

Answer 9

We have
\[(2xy^2+2y) + (2x^2 y + 2x) \frac{dy}{dx} = 0 \]
writing this as
\[(2xy^2+2y)dx+(2x^2y+2x)dy\]
If you notice the terms the first term is the derivative of \(x^2y^2+2xy\) with respect to x
and the second term is the derivative of \(x^2y^2+2xy\) with respect to y.
So this can be written as
\[d(x^2y^2+2xy)=0\]
you do know that
\[ d(xy)= ydx+xdy\]
and
\[d(x^2y^2)=2xy^2dx+2x^2ydy\]
therefore we have
\[d(x^2y^2+2xy)=0\]
integrate both sides, you'll get
\[x^2y^2+2xy=C\]