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romeo.nkala Group Title

hello guys

  • 2 years ago
  • 2 years ago

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  1. sunlove Group Title
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    we cant chat here this is a question only zone

    • 2 years ago
  2. sunlove Group Title
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    sorry

    • 2 years ago
  3. sunlove Group Title
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    we will get in trouble if we do

    • 2 years ago
  4. ash2326 Group Title
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    @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:)

    • 2 years ago
  5. romeo.nkala Group Title
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    ow i apologise...i was taught to greet, before i ask questions

    • 2 years ago
  6. ash2326 Group Title
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    No worries, just ask questions:)

    • 2 years ago
  7. romeo.nkala Group Title
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    2(y+3)dx-xydy=0 im asked to find differential equation

    • 2 years ago
  8. ash2326 Group Title
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    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

    • 2 years ago
  9. romeo.nkala Group Title
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    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

    • 2 years ago
  10. ash2326 Group Title
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    Hey romeo, I was not here. What you don't understand?

    • 2 years ago
  11. romeo.nkala Group Title
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    the whole point of de

    • 2 years ago
  12. ash2326 Group Title
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    Okay do you know differentiation and integration?

    • 2 years ago
  13. romeo.nkala Group Title
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    yep 100%

    • 2 years ago
  14. ash2326 Group Title
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    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?

    • 2 years ago
  15. romeo.nkala Group Title
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    yes that makes sense

    • 2 years ago
  16. ash2326 Group Title
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    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.

    • 2 years ago
  17. romeo.nkala Group Title
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    i see

    • 2 years ago
  18. ash2326 Group Title
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    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

    • 2 years ago
  19. romeo.nkala Group Title
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    wow ok.. so solving a de is solving for y?

    • 2 years ago
  20. ash2326 Group Title
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    Yeah

    • 2 years ago
  21. ash2326 Group Title
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    or any other dependent variable

    • 2 years ago
  22. ash2326 Group Title
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    like u=5v u is dependent and v is independent

    • 2 years ago
  23. romeo.nkala Group Title
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    so do u always have to rearrange such that its in the form f(y)dy=g(x)dx?

    • 2 years ago
  24. ash2326 Group Title
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    Yeah otherwise we can't integrate

    • 2 years ago
  25. romeo.nkala Group Title
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    can we try this one \[(x ^{2}- xy+y ^{2})dx-xydy=0\]

    • 2 years ago
  26. ash2326 Group Title
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    Okay :)

    • 2 years ago
  27. romeo.nkala Group Title
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    i cant seem to separate xy

    • 2 years ago
  28. ash2326 Group Title
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    It's a little complicated. There's no way it can be separated. it requires a different MO

    • 2 years ago
  29. ash2326 Group Title
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    MO or procedure. I'll explain you.

    • 2 years ago
  30. romeo.nkala Group Title
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    ok

    • 2 years ago
  31. ash2326 Group Title
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    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

    • 2 years ago
  32. romeo.nkala Group Title
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    i think so

    • 2 years ago
  33. ash2326 Group Title
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    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

    • 2 years ago
  34. ash2326 Group Title
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    DId you get it @romeo.nkala ?

    • 2 years ago
  35. romeo.nkala Group Title
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    yep got it. its the same

    • 2 years ago
  36. ash2326 Group Title
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    Yeah:D. This is just the basics of DE. You'll learn complex ones later:D

    • 2 years ago
  37. romeo.nkala Group Title
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    newayz! basic?

    • 2 years ago
  38. romeo.nkala Group Title
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    thanks alot!! by the way

    • 2 years ago
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