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romeo.nkala

  • 3 years ago

hello guys

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  1. sunlove
    • 3 years ago
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    we cant chat here this is a question only zone

  2. sunlove
    • 3 years ago
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    sorry

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

  4. ash2326
    • 3 years ago
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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:)

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

  6. ash2326
    • 3 years ago
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    No worries, just ask questions:)

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

  8. ash2326
    • 3 years ago
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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

  9. romeo.nkala
    • 3 years ago
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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

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

  11. romeo.nkala
    • 3 years ago
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    the whole point of de

  12. ash2326
    • 3 years ago
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    Okay do you know differentiation and integration?

  13. romeo.nkala
    • 3 years ago
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    yep 100%

  14. ash2326
    • 3 years ago
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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?

  15. romeo.nkala
    • 3 years ago
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    yes that makes sense

  16. ash2326
    • 3 years ago
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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.

  17. romeo.nkala
    • 3 years ago
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    i see

  18. ash2326
    • 3 years ago
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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

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

  20. ash2326
    • 3 years ago
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    Yeah

  21. ash2326
    • 3 years ago
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    or any other dependent variable

  22. ash2326
    • 3 years ago
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    like u=5v u is dependent and v is independent

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

  24. ash2326
    • 3 years ago
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    Yeah otherwise we can't integrate

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

  26. ash2326
    • 3 years ago
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    Okay :)

  27. romeo.nkala
    • 3 years ago
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    i cant seem to separate xy

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

  29. ash2326
    • 3 years ago
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    MO or procedure. I'll explain you.

  30. romeo.nkala
    • 3 years ago
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    ok

  31. ash2326
    • 3 years ago
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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

  32. romeo.nkala
    • 3 years ago
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    i think so

  33. ash2326
    • 3 years ago
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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

  34. ash2326
    • 3 years ago
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    DId you get it @romeo.nkala ?

  35. romeo.nkala
    • 3 years ago
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    yep got it. its the same

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

  37. romeo.nkala
    • 3 years ago
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    newayz! basic?

  38. romeo.nkala
    • 3 years ago
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    thanks alot!! by the way

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