1. anonymous

22xdy+11ydx+3ydy=0

2. anonymous

Loki! haha I'm back with more questions. :)

3. anonymous

Hehe, I see...

4. anonymous

Are you studying exact differential equations?

5. anonymous

I don't think so. I know what the answer is and its not a number. Its the formula of the constant

6. anonymous

Can you wait a little bit - I'm just finishing another post.

7. anonymous

Sure

8. anonymous

22xdy+11ydx+ydy=0 is the actual problem, Sorry

9. anonymous

ok

10. anonymous

i need paper

11. anonymous

haha alright

12. anonymous

oh, i don't have her question.

13. anonymous

Scotty, this looks worse than it is (sorry, I have someone else I'm helping here too).

14. anonymous

I'm hoping so. I think it fits into the trick of d(xy)=dy+ydx am I correct?

15. anonymous

Hang on...I misread it...I have to work something out. Sorry, I'll just finish with this other person and I won't be distracted.

16. anonymous

Ok so the problem is 22xdy+11ydx+3ydy=0

17. anonymous

I wish that 22 wasn't there. You sure it's not 11?

18. anonymous

Positive :/

19. anonymous

I know what the answer if, if that helps.

20. anonymous

yeah punch it in

21. anonymous

Its $11xy^2+y^3=C$

22. anonymous

Sorry, I keep being called away.

23. anonymous

It's alright..

24. anonymous

I'm thinking a substitution method. Just one sec...again called away...sorry. Won't be forever.

25. anonymous

I think I may have your method...just let me check properly.

26. anonymous

Ok cool

27. anonymous

I keep being called away! I'll just let you know what I'm thinking and you can play with it as well. Your equation, when rearranged, looks like this$y'=-\frac{11y}{22x+y}=-\frac{11(\frac{y}{x})}{22+(\frac{y}{x})}$after dividing the numerator and denominator by x.

28. anonymous

This is in the form for a type of differential equation that can be solved when we know that y' is some function of y/x; that is,$y'=F(\frac{y}{x})$

29. anonymous

We make the identification,$v(x)=\frac{y}{x}$so that$y=xv$

30. anonymous

and we take it from there. Is this familiar?

31. anonymous

Ok let me try to pmess with it

32. anonymous

Ok instead of that one... can you help me solve y'-9y=8e^x

33. anonymous

I'm doing this one for a timed quiz

34. anonymous

I figured out your other problem - I made an arithmetic error, but the method is correct.

35. anonymous

Yeah I'll do the second.

36. anonymous

Ok I think I saw where you were going with it and I gutted my way through it. Thanks

37. anonymous

Okay, integrating factor one - do you know?

38. anonymous

Not sure what that means..

39. anonymous

$y'-9y=8e^x \rightarrow \frac{d (\mu y)}{dx}=\mu 8e^x$

40. anonymous

$\mu = e^{\int\limits_{}{}-9dx}=e^{-9x}$

41. anonymous

so

42. anonymous

im sorry that should be y"

43. anonymous

$e^{-9x}y=e^{-9x}.8e^x+c \rightarrow y=ce^{9x}-e^x$

44. anonymous

y''-9y=8e^x ?

45. anonymous

Yes, sorry for the confusion. I miss typed.

46. anonymous

right do you just want the answer for the quiz and then the method

47. anonymous

Well. I want to learn but first and foremost I really need to pass this quiz

48. anonymous

I am writing it all down to try to understand though.

49. anonymous

$y=c_1e^{3x}+c_2e^{-3x}-e^{x}$

50. anonymous

I'll explain...

51. anonymous

Are you sure the + isn't a - between the first 2 terms?

52. anonymous

You have a second order differential equation which is linear, inhomogeneous (because the RHS is a function of x only) with constant coefficients. You need to solve inhom. equations by 1) solving the equivalent homogeneous equation first 2) finding a particular solution

53. anonymous

+

54. anonymous

In the end, it doesn't matter because it's absorbed into the constant.

55. anonymous

The only option I have with a + sign in it is... -2e^3x+e^-3x-e^x

56. anonymous

Ok so e^3x=e^-3x-e^x is the same thing?

57. anonymous

Oh, do you have initial conditions?

58. anonymous

Sorry when x=0

59. anonymous

Your equation has it's constants solve for

60. anonymous

Oh...you need to give me the whole question ;)

61. anonymous

Wait.... y=0 and y'=8 and x=0... I'm sorry I've been up all day doing this

62. anonymous

It's second order, so there should be 2 conds.

63. anonymous

ok

64. anonymous

c_1 = 2 and c_2 = -1

65. anonymous

$y=2e^3x-e^{-3x}-e^x$

66. anonymous

Yep that works!

67. anonymous

Good

68. anonymous

Thanks so much... I really feel low just asking for answers but I plan on taking the class again. If I fail I have to pay the whole tuition and I've been studying so hard and just not getting it lately :(

69. anonymous

It's because you're having to do it on your own - you're not being shown the ropes. Don't punish yourself.

70. anonymous

trying not to. Could you help me with 1 or 2 more? Then I'll leave you alone.

71. anonymous

yes

72. anonymous

How long do you have to do the quiz?

73. anonymous

i panic in tests - you shouldn't have told me, lol

74. anonymous

4xydx=(x^2+4)dy and that is the whole problem and not leaving anything out :)

75. anonymous

It's separable.

76. anonymous

$4xydx=(x^2+4)dy \rightarrow \frac{4xy}{x^2+4}=\frac{dy}{dx} \rightarrow \frac{4x}{x^2+4}dx=\frac{dy}{y}$

77. anonymous

You can integrate from there.

78. anonymous

Ok let me try

79. anonymous

$y=c(x^2+4)^2$

80. anonymous

yep thats what I got

81. anonymous

Aesome

82. anonymous

or, awesome

83. anonymous

Ok I think this should be the last one... I think I got the others actually.

84. anonymous

ok

85. anonymous

dy+2ydx=e^(-4x)dx

86. anonymous

It's an integrating factor one again.

87. anonymous

Yeah those are really difficult for me. Seperatable ones are the ones I'm more comfortable with...

88. anonymous

$y=-\frac{1}{2}e^{-4x}+ce^{-2x}$

89. anonymous

I'm just skipping to the solution because of the test...

90. anonymous

The closest answer I have that looks like that is e^(4x)+Ce^(2x)

91. anonymous

hmmm...that's the solution to the equation you gave me...

92. anonymous

let me write it in the formula so it looks better maybe it looks funny on text

93. anonymous

$dy+2ydx=e^{-4x}dx \rightarrow y'+2y=e^{-4x}$?

94. anonymous

$dy-2ydx=6e^(5x) dx$

95. anonymous

that dan 5x is super script

96. anonymous

so you gave me the wrong question before?

97. anonymous

Im a failure sorry I'm rushing

98. anonymous

$y(x) = c_1 e^{2 x}+2 e^{5 x}$

99. anonymous

is that one available?

100. anonymous

yes just not written that way but yeah

101. anonymous

I had actually guesed that one after trying to work it myself. Hurray for small miracles.

102. anonymous

so you've passed?

103. anonymous

Well... not 100% sure. I think I got the rest of them right.

104. anonymous

Do you still want to look at the first one?

105. anonymous

For dy+2ydx=e^(-4x)dx I got.... y=-1/2e^(-4x)+ce^(-2x)

106. anonymous

that's what I got..?

107. anonymous

and thats the only other one I kind of guessed on. I do want to look at those though yes.

108. anonymous

No I did that one myself

109. anonymous

good

110. anonymous

I think its right, not 100% sure though

111. anonymous

It's right

112. anonymous

if

113. anonymous

it's the one I did before...and it is...it's right.

114. anonymous

You're fine at this!

115. anonymous

Oh ok, sorry I'm scattered. ok I'm submitting it!

116. anonymous

117. anonymous

How much time do you have?

118. anonymous

5 mins left

119. anonymous

the first one was a practice but I do want to know how to do it

120. anonymous

ok

121. anonymous

122. anonymous

I said yes.

123. anonymous

yes, separable

124. anonymous

ok and 2xdx+ysinxdy=sinxdx

125. anonymous

I got y^2=x-2sinx+C

126. anonymous

that's separable too - is that how you solved it?

127. anonymous

Yes

128. anonymous

that one would have taken a bit of effort

129. anonymous

Ok is y'=y/x y=cx

130. anonymous

yes

131. anonymous

ok well it looks all good then :)

132. anonymous

Thank you so much for that man.....

133. anonymous

that's okay - i hope you pass

134. anonymous

have you heard of a site called wolframalpha? www.wolframalpha.com

135. anonymous

Ok so that original problem..

136. anonymous

You can type in things and it will calculate, like your diff. eqs

137. anonymous

but it won't show you what to do, that's the only thing

138. anonymous

I have, I don't know how to type these kinds of problems in though.... could you show me how?

139. anonymous

I'm just thinking, if you have more of these tests, you can have a look at that as a resource

140. anonymous

I did use it before, I have 2 more tests actually.

141. anonymous

y'-2y=e^(-4x) is what you can type in for one of your d.e.'s above. It will give you the answer.

142. anonymous

Use primes on the y's...

143. anonymous

You don't have to tell it to solve for y; it guesses. If it doesn't, then it's a form it doesn't recognize.

144. anonymous

Hmm.... but even for stuff like... 22xdy+11ydx+3ydy=0?

145. anonymous

Yeah, divide everything on your paper by dx so you can have (dy/dx) which you can equate as y'

146. anonymous

lol... thats a genius way to do it.

147. anonymous

$22x \frac{dy}{dx}+y+3y \frac{dy}{dx}=0 \rightarrow 22xy'+y+3yy'=0$

148. anonymous

I don't know why they keep giving you the equation in that form...it's weird.

149. anonymous

In the form I gave you? Thats the way my teacher gives me them....

150. anonymous

Yeah, I know...weird.

151. anonymous

Almost all my questions are like that is that not how they are normally done?

152. anonymous

lol, no

153. anonymous

Hmm why would she try to teach like that then?

154. anonymous

Maybe she's getting you ready to think in terms of differentials or something. Really, it doesn't matter. Just stick to what she does.

155. anonymous

did you pass this quiz?

156. anonymous

Ok well I really appreciate it. You might be the only reason I passed that quiz... I'm sure of it.

157. anonymous

I won't know for a day or two, but I'm sure I did.

158. anonymous

Cool, hope so...don't want to be spending more money on tuition...

159. anonymous

god I know. I have...2 more quizes and a final. Then I'll be done with this class.

160. anonymous

Good luck then. If you need help, look up an old post and send something. I'll get an e-mail about it.

161. anonymous

Ok I will man. thanks. Hopefully if you're around I can ask for your help like this again I would star you 1000 times if I could

162. anonymous

No worries. have a go at that first one. The method of substitution v=y/x should work. You end up with a differential equation in v which is separable...and you're comfortable with that ;)

163. anonymous

I'm going to go offline for a while. I have some stuff to do. Good luck!

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