anonymous
  • anonymous
hello guys
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
we cant chat here this is a question only zone
anonymous
  • anonymous
sorry
anonymous
  • anonymous
we will get in trouble if we do

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ash2326
  • 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
  • anonymous
ow i apologise...i was taught to greet, before i ask questions
ash2326
  • ash2326
No worries, just ask questions:)
anonymous
  • anonymous
2(y+3)dx-xydy=0 im asked to find differential equation
ash2326
  • 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
  • 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
  • ash2326
Hey romeo, I was not here. What you don't understand?
anonymous
  • anonymous
the whole point of de
ash2326
  • ash2326
Okay do you know differentiation and integration?
anonymous
  • anonymous
yep 100%
ash2326
  • 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
  • anonymous
yes that makes sense
ash2326
  • 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
  • anonymous
i see
ash2326
  • 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
  • anonymous
wow ok.. so solving a de is solving for y?
ash2326
  • ash2326
Yeah
ash2326
  • ash2326
or any other dependent variable
ash2326
  • ash2326
like u=5v u is dependent and v is independent
anonymous
  • anonymous
so do u always have to rearrange such that its in the form f(y)dy=g(x)dx?
ash2326
  • ash2326
Yeah otherwise we can't integrate
anonymous
  • anonymous
can we try this one \[(x ^{2}- xy+y ^{2})dx-xydy=0\]
ash2326
  • ash2326
Okay :)
anonymous
  • anonymous
i cant seem to separate xy
ash2326
  • ash2326
It's a little complicated. There's no way it can be separated. it requires a different MO
ash2326
  • ash2326
MO or procedure. I'll explain you.
anonymous
  • anonymous
ok
ash2326
  • 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
  • anonymous
i think so
ash2326
  • 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
  • ash2326
DId you get it @romeo.nkala ?
anonymous
  • anonymous
yep got it. its the same
ash2326
  • ash2326
Yeah:D. This is just the basics of DE. You'll learn complex ones later:D
anonymous
  • anonymous
newayz! basic?
anonymous
  • anonymous
thanks alot!! by the way

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