## anonymous 3 years ago answered question

1. anonymous

bernoulli's Equation

2. UnkleRhaukus

what do you get if you re arrange to dy/dx=

3. anonymous

sorry but i dont know .................it equal to 1 dont know if I am correct >.<

4. UnkleRhaukus

after your get the left hand side to be dy/dx sub v=y/x

5. anonymous

$\frac{ xdy }{ y-xdx }=1$

6. anonymous

its right?

7. UnkleRhaukus

yeah, now move the other terms to the other side

8. anonymous

$\frac{ dy }{ dx }= \frac{ y-x }{ x }$

9. anonymous

Im correct?

10. UnkleRhaukus

good

11. anonymous

then?

12. anonymous

let v

13. anonymous

how?

14. UnkleRhaukus

$(y-x)/x = y/x - x/x$

15. anonymous

yes

16. anonymous

after that

17. UnkleRhaukus

x/x simplifies to ...

18. anonymous

1

19. UnkleRhaukus

good so you have $\frac{dy}{dx}=\frac yx-1$ now let $$y/x=v$$ $$y=vx$$ $$y'=?$$

20. UnkleRhaukus

use the product rule to find y'

21. anonymous

$y'=v(1) -x\frac{dv }{ dx }$

22. UnkleRhaukus

should be plus not minus

23. anonymous

24. anonymous

y′=v(1)+xdv/dx

25. UnkleRhaukus

now sub this in $\frac{dy}{dx}=y/x-1$ becomes $v+x\frac{dv}{dx}=v-1$

26. UnkleRhaukus

which is now variables separable

27. anonymous

$\frac{ dv }{ dx }= \frac{ v ^{2} -v}{ x }$

28. UnkleRhaukus

um you should have taken away the v from both sides, before dividing by x

29. anonymous

$$(x-y)\,dx+x\,dy=0$$note this is homogeneous; both our functions $$x-y,x$$ are homogeneous. as @UnkleRhaukus stated, use a substitution $$y=vx$$ i.e. $$dy=\left(x\dfrac{dv}{dx}+v\right)\,dx$$:(x-vx)dx+x(x\,dv+v\,dx)=0\$$x-vx)\,dx+x^2dv+vx\,dx=0\\x\,dx+x^2\,dv=0\\x^2\,dv=-x\,dx\\dv=-\frac1x\,dx\\v=-\log x+C\\y/x=-\log x+C\\y=-x\log x+Cx 30. anonymous 31. anonymous guys can be this one $xdv=(v ^{2}-v) dx .....then....... (v ^{2}-v)-xdv=0$ 32. anonymous ? 33. anonymous then identify if its exact DE or not 34. anonymous ????? 35. UnkleRhaukus $v+x\frac{dv}{dx}=v-1\\ x\frac{dv}{dx}=-1\\dv=-xdx\\ \\∫dv=-∫dx/x$ 36. UnkleRhaukus What is the second thing? where does the dx go? 37. anonymous v=- lnx +c is that right? 38. UnkleRhaukus yes, now remember v=y/x 39. anonymous then sub v= y/x to v =-ln x +c 40. anonymous y = -xlnx+c ? 41. UnkleRhaukus almost, 42. anonymous how ? 43. UnkleRhaukus the last term, 44. anonymous cx 45. anonymous yes 46. anonymous sorry about that 47. UnkleRhaukus thats better \(\checkmark$$

48. anonymous

tnx i other one here tech me ?

49. anonymous

xdx + (y-2x )dy

50. UnkleRhaukus

is there an equals sign somewhere?

51. anonymous

=0

52. anonymous

$\frac{ dy }{ dx }=\frac{ -x }{ y- }$

53. anonymous

2x

54. UnkleRhaukus

ok this one is the same technique, rearrange to dy/dx= and substitute v=y/x

55. UnkleRhaukus

divide both sides of the fraction by x, to get it in the right form $\frac{-x}{y-2x}=\frac{-x/x}{y/x-2x/x }$

56. anonymous

always? v=y/x even to other problem?

57. UnkleRhaukus

When we can express the DE as y'(x) = f (y/x) we can call this a homogenous equation and v = y/x substitution will work if we can't get this form y'(x) = f (y/x) this technique wont work

58. anonymous

ah i see

59. anonymous

can you gve to me the final answer here this will serve as my basis

60. UnkleRhaukus

sometimes a 'homogenous equations" means something else

61. anonymous

you gve a sense to me :)

62. UnkleRhaukus

i haven't worked out the final answer yet

63. anonymous

by the way thank you :))) tnz alot