## MathSofiya Group Title Solve using variation of parameters $y''-y'=e^x$ $r^2-r=0$ $r_1=r_2=1$ $y_c=c_1e^x+c_2xe^x$ $W(y_1,y_2)=0$ Is that possible? one year ago one year ago

1. MathSofiya Group Title

No I did that wrong, one moment

2. MathSofiya Group Title

no it equals $e^{2x}$

3. MathSofiya Group Title

$y_p(x)=-e^xx^2+x^2e^x$

4. TuringTest Group Title

wrong complimentary

5. TuringTest Group Title

$r^2-r=r(r-1)=0\implies r=\{?,?\}$

6. MathSofiya Group Title

$r^2-r=0$

7. MathSofiya Group Title

r=0 and r=1

8. TuringTest Group Title

so the complimentary is...?

9. MathSofiya Group Title

$y_c=c_1+c_2e^x$

10. TuringTest Group Title

right, now I get a bit dicey.... since you have e^x on the RHS as well I feel you may need to multiply this by x to keep it linearly independent

11. MathSofiya Group Title

sure $y_c=c_1+c_2xe^x$

12. TuringTest Group Title

$y_c=c_1x+c_2xe^x$

13. MathSofiya Group Title

why on both c_1 and c_2, how did come to that conclusion?

14. TuringTest Group Title

or do you do it to the particular instead? hm... well normally you do it to the whole RHS particular, but since you need the complimentary for the particular in VP I'm not sure...

15. MathSofiya Group Title

I'm not sure... DunnBros is about to kick me out, they close at 11pm

16. MathSofiya Group Title

I'll be back on in like 15 mins when I get home

17. TuringTest Group Title

fair enough, I'm about to call it a night as well see ya, I will investigate this :)

18. MathSofiya Group Title

Make that 7 mins, ha!

19. MathSofiya Group Title

$W(y_1,y_2)=x^2e^x$

20. MathSofiya Group Title

$y_1=x$ $y_2=xe^x$

21. MathSofiya Group Title

$y_p(x)=-x\int \frac{e^x}xdx+xe^x\int x^{-1}dx$

22. wasiqss Group Title

well ik it

23. wasiqss Group Title

my twin :D

24. MathSofiya Group Title

hi twin

25. TuringTest Group Title

that first integral is not going to be so nice I don't think.... you know who knows exactly what we're missing @lalaly DE help please!!!!!

26. wasiqss Group Title

well i can do it

27. wasiqss Group Title

tell me equation for which i have to solve\y complimentary

28. MathSofiya Group Title

y''-y'=e^x

29. TuringTest Group Title

we need variation of parameters @wasiqss

30. wasiqss Group Title

yehh just tell me equation i ll take out solution. as y complimentary is solved i ll solve for y particular

31. TuringTest Group Title

Our problem is that the solutions to the complimentary are not independent from the particular, which you can tell since we got W=0 on our first try

32. TuringTest Group Title

ok @wasiqss let's see what you got :)

33. wasiqss Group Title

tell me equation man completely cause this seems messy

34. TuringTest Group Title

y''-y'=e^x

35. wasiqss Group Title

easy one lol

36. TuringTest Group Title

then do it, by all mean!

37. MathSofiya Group Title
38. wasiqss Group Title

y complimentary is y= c1 +c2e^-x

39. TuringTest Group Title

^yes, answer not important, we need the process...

40. TuringTest Group Title

complimentary is y=c1+c2e^x

41. MathSofiya Group Title

i disagree with the above answer, shouldn't it be e^x, yeah what turing has

42. wasiqss Group Title

yes cause roots are zero and -1

43. TuringTest Group Title

no they are not

44. TuringTest Group Title

r^2-r=0 r(r-1)=0 r={0,1}

45. MathSofiya Group Title

do you think my W is right? x^2 e^x

46. TuringTest Group Title

that I doube @MathSofiya because we have e^x on both left and right our solutions are linearly dependet (that means we can't solve it yet) so maybe if we multiply the g(x) by x ? not sure, but worth a try

47. TuringTest Group Title

with W=e^2e^x I don't think that first integral is closed, check the wolf, I have not yet

48. TuringTest Group Title

w=x^2e^x *

49. MathSofiya Group Title

do we agree that $y_c=c_1x+c_2xe^x$ or should it be $y_c=c_1+c_2e^x$

50. MathSofiya Group Title

51. TuringTest Group Title

that I am not sure, but based on our attempt to use the second one failing I would now try the first (we are discovering this together right now :)

52. TuringTest Group Title

however if we are to try the first we need to multiply g(x) by x (I'm thinking....)

53. TuringTest Group Title

so let's try that

54. TuringTest Group Title

otherwise we get W=0 which is a failure

55. TuringTest Group Title

@lalaly I know you can help us, please where are you?

56. MathSofiya Group Title

$y_c=c_1x+c_2xe^x$ $\left|\begin{matrix}x&xe^x\\1&e^x(x-1)\end{matrix}\right|=x^2e^x$

57. MathSofiya Group Title

can you explain the g(x) concept again and why it being e^x would be a problem because that cancels something out?...

58. TuringTest Group Title

no I think that is wrong because that leads to the particular having$\int\frac{e^x}dxx$

59. TuringTest Group Title

*$\int\frac{e^x}xdx$

60. TuringTest Group Title

here is the problem... have you taken linear algebra?

61. MathSofiya Group Title

no

62. MathSofiya Group Title

@lalaly is here

63. TuringTest Group Title

okay. then we have a set of vectors (in this case equations) if their determinant is zero we know they are linearly dependent

64. wasiqss Group Title

now listen

65. wasiqss Group Title

for y particular

66. TuringTest Group Title

that is a bad thing in DE's we need all of our solutions to be linearly independent, so the determinant of the solutions, (the wronskian) must not be zero, otherwise our solutions are linearly dependent

67. wasiqss Group Title

is it xe^x

68. wasiqss Group Title

so as the denominator becomes zero

69. TuringTest Group Title

and would you care to explain how/why ?

70. wasiqss Group Title

we multiply by x

71. wasiqss Group Title

cofficeint of e^1x

72. wasiqss Group Title

=1

73. wasiqss Group Title

[xe^]/[(-1)^2-1)]

74. wasiqss Group Title

HENCE we get zero in denominator

75. TuringTest Group Title

anyway... to keep the particular linearly independent from the complimentary we often need to multiply by x, so that is what I am thinking must happen somehow @lalaly will confirm if I am making any sense or not now

76. wasiqss Group Title

now we need to apply binomeal expansion

77. TuringTest Group Title

@wasiqss sorry to be so direct, but no we do not need any binomial expansion here please let's just have a listen to @lalaly

78. wasiqss Group Title

arghh i wish i was good at making you understand it

79. wasiqss Group Title

we need it!

80. TuringTest Group Title

you cannot have 0 in the denom, that shows you that you have already made a mistake !!!!

81. wasiqss Group Title

well to counter this zero!! we need to multiply x

82. TuringTest Group Title

...as I was saying...

83. wasiqss Group Title

plz no offense but plz go revise DE s

84. TuringTest Group Title

._.

85. wasiqss Group Title

you know what .. we can approach a DE in differnt ways and this one i will certainly approach by binomeal.. cause we studied this way for this case

86. TuringTest Group Title

yes, but I think it is way more work than is necessary now I just wanna hear @lalaly

87. wasiqss Group Title

kay

88. TuringTest Group Title

oh dear, we seem to have lost her I am going to try this op paper I guess

89. wasiqss Group Title

hahaha lana dear

90. wasiqss Group Title

$[1+(D(D+1)-1)^-1[x^2e^x$

91. wasiqss Group Title

we have to expand binomeally [1+(D(D+1)−1)−1

92. wasiqss Group Title

well i have a solution on page i wonder how to give you that

93. MathSofiya Group Title

take a picture with your cellphone and attach it

94. MathSofiya Group Title

or use $\LaTeX$

95. wasiqss Group Title

good idea twin can i do it tomorow because i would need to upload it too?

96. TuringTest Group Title

okay got it :)

97. wasiqss Group Title

latex is something not my type

98. wasiqss Group Title

turing my approach?

99. TuringTest Group Title

do everything normally, then multiply the particular by x at the end that is all

100. MathSofiya Group Title

get used to it, it's a very helpful Latex is a very useful tool

101. wasiqss Group Title

turing binomeal is essential in it!

102. TuringTest Group Title

oh yeah?, just watch...

103. TuringTest Group Title

actually no need to multiply by x at all$y_c=c_1+c_2e^x$$W=e^x$$y_p=-\int e^{-x}+e^x\int dx=e^x+xe^x$$y_p=e^x+xe^x$$y=y_c+y_p=c_1+c_2e^x+e^x+xe^x=c_1+e^x(1+x+c_2)$and yes, but is your solution in 5 lines?

104. TuringTest Group Title

oh crap I see a mistake

105. wasiqss Group Title

wait you told me w=xe^x!

106. MathSofiya Group Title

dude this whole discussion was based on finding what W equaled

107. wasiqss Group Title

108. TuringTest Group Title

@MathSofiya yes, because the entire problem we were having was keeping our solutions linearly independent, which we check by finding W

109. wasiqss Group Title

waittttttttttttt what the....... i was solving by another method not by variation of parameter lol

110. TuringTest Group Title

$y_c=c_1+c_2e^x$$W=e^x$$y_p=-\int e ^x dx+e^x\int dx=-e^x+xe^x$$y=y_c+y_p=c_1+c_2e^x-e^x+xe^x=c_1+e^x(x-1+c_2)$

111. TuringTest Group Title

there we go... stupid algebra tripped me up :/

112. MathSofiya Group Title

yeah that's what I get $-e^x+e^xx$

113. TuringTest Group Title

so yc+yp gives what I have, and what wolf has so we're good :)

114. MathSofiya Group Title

good. Hi 5

115. TuringTest Group Title

back atchya!

116. MathSofiya Group Title

As always thank you!

117. TuringTest Group Title

As always you're welcome :D