hello guys

- anonymous

hello guys

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

we cant chat here this is a question only zone

- anonymous

sorry

- anonymous

we will get in trouble if we do

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- ash2326

@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:)

- anonymous

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

- ash2326

No worries, just ask questions:)

- anonymous

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

- ash2326

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

- anonymous

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

- ash2326

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

- anonymous

the whole point of de

- ash2326

Okay do you know differentiation and integration?

- anonymous

yep 100%

- ash2326

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?

- anonymous

yes that makes sense

- ash2326

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.

- anonymous

i see

- ash2326

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

- anonymous

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

- ash2326

Yeah

- ash2326

or any other dependent variable

- ash2326

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

- anonymous

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

- ash2326

Yeah otherwise we can't integrate

- anonymous

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

- ash2326

Okay
:)

- anonymous

i cant seem to separate xy

- ash2326

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

- ash2326

MO or procedure. I'll explain you.

- anonymous

ok

- ash2326

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

- anonymous

i think so

- ash2326

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

- ash2326

DId you get it @romeo.nkala ?

- anonymous

yep got it. its the same

- ash2326

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

- anonymous

newayz! basic?

- anonymous

thanks alot!! by the way

Looking for something else?

Not the answer you are looking for? Search for more explanations.