Question

hello guys

Answer 1

we cant chat here this is a question only zone

Answer 2

sorry

Answer 3

we will get in trouble if we do

Answer 4

@romeo.nkala Welcome to Open Study!!!!
If you need to talk to someone use chat, don't post it here. This is for questions only
Thanks:)

Answer 5

ow i apologise...i was taught to greet, before i ask questions

Answer 6

No worries, just ask questions:)

Answer 7

2(y+3)dx-xydy=0 im asked to find differential equation

Answer 8

We have
\[2(y+3)dx=xydy\]
Let's bring x to one side and y to other
we get
\[2\frac{dx}{x}=\frac{y}{y+3} dy\]
Now we'll integrate both sides
\[\int 2\frac{dx}{x}=\int \frac{y+3-3}{y+3} dy\]
\[\int 2\frac{dx}{x}= \int (1-\frac{3}{y+3}) dy\]
Now integrating
we get
\[2 \ln x+C= y- 3 \ln {(y+3)}\]
or
\[y=2 \ln x-3\ln (y+3) +C\]
C= constant of integration

Answer 9

im doing this for the 1st time, i read through the note i cant seem to get the whole point and procedure. can you explain to me the important stuff

Answer 10

Hey romeo, I was not here. What you don't understand?

Answer 11

the whole point of de

Answer 12

Okay do you know differentiation and integration?

Answer 13

yep 100%

Answer 14

Suppose we have \[
y=x^2\]
Let's differentiate this
we get
\[\frac{dy}{dx}=2x\]
so
\[\frac{dy}{dx}-2x=0\]
This is a differential equation, it can also be written as
\[dy-2xdx=0
\]
Here we started with y=x^2, that's why we know that the solution of the differential equation is y=x^2
Did you understand this?

Answer 15

yes that makes sense

Answer 16

Here we had the solution from that we found the differential equation but it's not the case every time. We'll have the DE and we'd be required to find its solution.

Answer 17

i see

Answer 18

In our question we are given
\[2(y+3)dx-xydy=0 \]
We have to integrate this to find the solution or y
If we have to solve a DE, we need it in this form
\[f(y)dy=g(x) dx\]
f is any function of y and g could be any function of x
What we see that with dy we need a function of y and with dx we need a function of x. We'd not like function of y with dx or vice-versa

Answer 19

wow ok.. so solving a de is solving for y?

Answer 20

Yeah

Answer 21

or any other dependent variable

Answer 22

like u=5v
u is dependent and v is independent

Answer 23

so do u always have to rearrange such that its in the form f(y)dy=g(x)dx?

Answer 24

Yeah otherwise we can't integrate

Answer 25

can we try this one \[(x ^{2}- xy+y ^{2})dx-xydy=0\]

Answer 26

Okay
:)

Answer 27

i cant seem to separate xy

Answer 28

It's a little complicated. There's no way it can be separated. it requires a different MO

Answer 29

MO or procedure. I'll explain you.

Answer 30

ok

Answer 31

We have
\[(x^2--xy+y^2)dx=xy dy\]
or
\[xydy=(x^2-xy+y^2) \]
Let's divide both sides by xy, we get
\[dy=(\frac{x}{y}-1+\frac{y}{x})dx\]
or
\[\frac{dy}{dx}=\frac{x}{y}-1+\frac{y}{x}\]
Now we'll substitute
\[v=\frac{y}{x}\]
or \[y=vx\]
Let's differentiate this we get
\[\frac{dy}{dx}=v+x\frac{dv}{dx}\]
Now substituting this in our DE, we get
\[v+x\frac{dv}{dx}=\frac{1}{v}-1+v\]
Can you separate v and x now?

V is the independent variable

Answer 32

i think so

Answer 33

Yeah now we'll get
\[x\frac{dv}{dx}=\frac{1-v}{v}\]
or
\[\frac{v}{1-v} dv=\frac{dx}{x}\]
Now you integrate both sides to find v, in terms of x and then substitute v=y/x to get your final solution

Answer 34

DId you get it @romeo.nkala ?

Answer 35

yep got it. its the same

Answer 36

Yeah:D. This is just the basics of DE. You'll learn complex ones later:D

Answer 37

newayz! basic?

Answer 38

thanks alot!! by the way