## romeo.nkala 3 years ago hello guys

1. sunlove

we cant chat here this is a question only zone

2. sunlove

sorry

3. sunlove

we will get in trouble if we do

4. 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:)

5. romeo.nkala

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

6. ash2326

7. romeo.nkala

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

8. 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

9. romeo.nkala

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

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

11. romeo.nkala

the whole point of de

12. ash2326

Okay do you know differentiation and integration?

13. romeo.nkala

yep 100%

14. 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?

15. romeo.nkala

yes that makes sense

16. 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.

17. romeo.nkala

i see

18. 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

19. romeo.nkala

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

20. ash2326

Yeah

21. ash2326

or any other dependent variable

22. ash2326

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

23. romeo.nkala

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

24. ash2326

Yeah otherwise we can't integrate

25. romeo.nkala

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

26. ash2326

Okay :)

27. romeo.nkala

i cant seem to separate xy

28. ash2326

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

29. ash2326

MO or procedure. I'll explain you.

30. romeo.nkala

ok

31. 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

32. romeo.nkala

i think so

33. 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

34. ash2326

DId you get it @romeo.nkala ?

35. romeo.nkala

yep got it. its the same

36. ash2326

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

37. romeo.nkala

newayz! basic?

38. romeo.nkala

thanks alot!! by the way