## melmel 2 years ago answered question

1. melmel

bernoulli's Equation

2. UnkleRhaukus

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

3. melmel

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. melmel

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

6. melmel

its right?

7. UnkleRhaukus

yeah, now move the other terms to the other side

8. melmel

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

9. melmel

Im correct?

10. UnkleRhaukus

good

11. melmel

then?

12. melmel

let v

13. melmel

how?

14. UnkleRhaukus

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

15. melmel

yes

16. melmel

after that

17. UnkleRhaukus

x/x simplifies to ...

18. melmel

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. melmel

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

22. UnkleRhaukus

should be plus not minus

23. melmel

24. melmel

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. melmel

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

28. UnkleRhaukus

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

29. oldrin.bataku

$$(x-y)\,dx+x\,dy=0$$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. oldrin.bataku 31. melmel guys can be this one $xdv=(v ^{2}-v) dx .....then....... (v ^{2}-v)-xdv=0$ 32. melmel ? 33. melmel then identify if its exact DE or not 34. melmel ????? 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. melmel v=- lnx +c is that right? 38. UnkleRhaukus yes, now remember v=y/x 39. melmel then sub v= y/x to v =-ln x +c 40. melmel y = -xlnx+c ? 41. UnkleRhaukus almost, 42. melmel how ? 43. UnkleRhaukus the last term, 44. melmel cx 45. oldrin.bataku yes 46. melmel sorry about that 47. UnkleRhaukus thats better \(\checkmark$$

48. melmel

tnx i other one here tech me ?

49. melmel

xdx + (y-2x )dy

50. UnkleRhaukus

is there an equals sign somewhere?

51. melmel

=0

52. melmel

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

53. melmel

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. melmel

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. melmel

ah i see

59. melmel

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. melmel

you gve a sense to me :)

62. UnkleRhaukus

i haven't worked out the final answer yet

63. melmel

by the way thank you :))) tnz alot