hello guys

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we cant chat here this is a question only zone
sorry
we will get in trouble if we do

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Other answers:

@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:)
ow i apologise...i was taught to greet, before i ask questions
No worries, just ask questions:)
2(y+3)dx-xydy=0 im asked to find differential equation
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
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
Hey romeo, I was not here. What you don't understand?
the whole point of de
Okay do you know differentiation and integration?
yep 100%
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?
yes that makes sense
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.
i see
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
wow ok.. so solving a de is solving for y?
Yeah
or any other dependent variable
like u=5v u is dependent and v is independent
so do u always have to rearrange such that its in the form f(y)dy=g(x)dx?
Yeah otherwise we can't integrate
can we try this one \[(x ^{2}- xy+y ^{2})dx-xydy=0\]
Okay :)
i cant seem to separate xy
It's a little complicated. There's no way it can be separated. it requires a different MO
MO or procedure. I'll explain you.
ok
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
i think so
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
DId you get it @romeo.nkala ?
yep got it. its the same
Yeah:D. This is just the basics of DE. You'll learn complex ones later:D
newayz! basic?
thanks alot!! by the way

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