## anonymous 5 years ago Show how to find the particular solution of the differential equation, dy/dx+5y=9, x=0 when y=1

1. anonymous

I could use a LaPlace transform right?

2. anonymous

Yeah.. and it's pretty easy.

3. anonymous

Ok the only problem I'm having is, all I have to use is the formula L(y')=sL(y)-y(0) correct?

4. anonymous

Yeah.

5. anonymous

ok so I'm having trouble figuring out how to incorperate x=0 and y=1

6. anonymous

x=0 when y=1 means that y(0)=1

7. anonymous

Oh.. duh.

8. anonymous

so I would just rewrite... $s L(y)-y(0)+5L(y)=L(9)$

9. anonymous

OK.. what are the L(y) and L(9)?

10. anonymous

Not 100% sure on L(y) but I know L(9)=9/s

11. anonymous

Would it just be y/s ?

12. anonymous

or no, I could solve for L(y)

13. anonymous

I think I have it

14. anonymous

Yeah.. you could do so, or you could write it as Y(s)

15. anonymous

You know that you have to transform it back to the time domain.

16. anonymous

L(y)=(9/s+1)/(s+5)

17. anonymous

are you sure about the simplification you made?

18. anonymous

oh I see it.. it's right.

19. anonymous

I don't know how to do fractions notation using the equation thing or I would. :)

20. anonymous

I think I need to grab a pen and a paper!

21. anonymous

Well you so far have: $Y(s)={9/s+1 \over s+5}$ you can multiply by s (since it's not 0).. you will get: ${9+s \over s(s+5)}$

22. anonymous

Oh right to clean it up. Ok so now how do I evaluate for the particular solution?

23. anonymous

${9+s \over s(s+5)}= {A \over s}+ {B \over s+5}$ how are you with partial fractions, can you find A and B?

24. anonymous

not good at ALL with partial fractions...

25. anonymous

it's easy.. multiply both sides of the last equation by s(s+5)

26. anonymous

You mean A(s(s+5)/s+B(s(s+5)/(s+5) ?

27. anonymous

Almost there!! something should be cancelled, right? And din't forget the left side of the equality sign.

28. anonymous

don't*

29. anonymous

A(s+5)+B(s)

30. anonymous

err

31. anonymous

that should equal to?

32. anonymous

9+s/s(s+5)=A(s+5)+B(s)

33. anonymous

nearly

34. anonymous

you multiply both sides by s(s+5)

35. anonymous

Oh right ok so its

36. anonymous

9+s=A(s+5)+B(s)

37. anonymous

Hey lokisan :).. sorry I took over. we seem to be useless wjen you're around :P

38. anonymous

you're doing fine :D

39. anonymous

I'm 'half' watching...

40. anonymous

there is more than one way find A and B now.. take s to be zero and find the value of A.

41. anonymous

If I did something wrong, Please let me know :)

42. anonymous

you're on the right track

43. anonymous

A=9/5

44. anonymous

Then I put that back into the original for B?

45. anonymous

or A and B for s...

46. anonymous

Yeah.. or just plug s=-5

47. anonymous

b=-4/5

48. anonymous

now what does Y(s) equal to?

49. anonymous

I'm kind of confused.. so I now put int he values of A and B back into 9+s=A(s+5)+B(s) to solve for the s?

50. anonymous

Do you know why we used the partial fractions in the first place?

51. anonymous

So we could evaluate what s was when x=0 and y=1

52. anonymous

No.. because we want to write an expression for Y(s) that can be easily transformed back to our original function.

53. anonymous

don't forget we are looking for the particular solution of the DE.

54. anonymous

My instructor never used the notation as Y(s) so I think thats where I'm losing you.

55. anonymous

Oh oh I think I understand... we're looking to change 9+s/s(s+5)

56. anonymous

$L(y)={9+5 \over s(s+5)}={9 \over 5}.{1 \over s}-{4 \over 5}.{1 \over s+5}$

57. anonymous

Isn't it 9+s

58. anonymous

oh yeah 9+s sorry

59. anonymous

now focus on the right part, and try to transform it back to y(x).

60. anonymous

Ugh I understand what you mean, I just am unaware of the method. I don't do inverse laplace right?

61. anonymous

You do nothing but the inverse Laplace.

62. anonymous

So like... 1/s=1 etc?

63. anonymous

Yeah.

64. anonymous

not 1/s=1.. but the inverse laplace of 1/s=1 or more accurately u(x), (which is the unit step function)

65. anonymous

How do I solve the 9/5 and -4/5

66. anonymous

laplace transform and its inverse are both linear operations, that's L[af(t)]=aL[f(t)]

67. anonymous

Ugh I was understanding it and now I'm lost. I understand to use the table but this is just beyond me now.

68. anonymous

You should not worry about these constants.. just keep them as they are.

69. anonymous

Am I complicating things here? :(

70. anonymous

So just worry about the 1/s and 1/s+5?

71. anonymous

No you're doing good. I'm sure its just me.

72. anonymous

yeah.

73. anonymous

you already did 1/s.. what about 1/(s+5)?

74. anonymous

I just think you're not familiar with notation.

75. anonymous

e^-at is the transform...

76. anonymous

e^-5t then right?

77. anonymous

You know what you need to do, though.

78. anonymous

Yes

79. anonymous

Me?

80. anonymous

No, Scotty.

81. anonymous

Exactly Scotty.

82. anonymous

Sorry...

83. anonymous

It's OK :)

84. anonymous

Ok so to clean it up I use use what I have left right?

85. anonymous

Which basically will end up e^5t

86. anonymous

How?

87. anonymous

Your answer should have e^(-5t) in it.

88. anonymous

9/5*1-4/5*e^(-5t)

89. anonymous

It's in x not in t.

90. anonymous

Yes, Scotty.

91. anonymous

But in x.

92. anonymous

I'm used to doing them in t...sorry.

93. anonymous

so e^-(5x)

94. anonymous

so final answer is: $y(x)={9 \over 5}- {4 \over 5} e ^{-5x}$ right?

95. anonymous

We all are used to t.

96. anonymous

perfect

97. anonymous

Ok I was rushing through on the simplification.

98. anonymous

I'm 12 years old and what is this? $\frac{\mathbb{d}y}{\mathbb{d}x} = 9 - 5y \implies \int \frac{1}{9-5y} \mathbb{d}y = \int \mathbb{d}x \implies c -\frac{1}{5} \ln 9 - 5y = x$ x = 0 when y = 1 $\implies c = \frac{1}{5} \ln 4 \implies x = \ln \frac{4}{9-5y} \implies y = \frac{9-4e^{-5x}}{5}$

99. anonymous

What?

100. anonymous

He needed to use Laplace.

101. anonymous

So that just gave me the general solution right? Now I evaluate the particular?

102. anonymous

He solved the problem without using the Laplace transform.

103. anonymous

Just kidding guys, no need to bully me :(

104. anonymous

LOL

105. anonymous

FWIW, he said ' I could us..', implying it wasn't necessary. And it isn't.

106. anonymous

Hey Micheal!!

107. anonymous

oh man...

108. anonymous

You spelt my name wrong. It's 'Isaac Newton'

109. anonymous

sure...

110. anonymous

Oh yeah?

111. anonymous

Im confused on what just happened.

112. anonymous

No. But you did spell my real name wrong.

113. anonymous

:(

114. anonymous

Sorry Michael!

115. anonymous

:D

116. anonymous

...

117. anonymous

So you just solved my problem doing tradidional integration then? Is that what I gather?

118. anonymous

Yeah.. He used separating variables method to solve the DE.

119. anonymous

You could have done it either way. You were on about Laplace, so Anwar answered as such.

120. anonymous

^^ Thank you!! You saved me :)

121. anonymous

I see, its not solved all the way though right? I have C and X, I just need to use those to solve y?

122. anonymous

Is this a differential equation course or what exactly?

123. anonymous

Its a Calc II with LaPlace and Diff Eq in it.

124. anonymous

A pain in the retriceonline course to be exact, where my teacher refuses to help me which is why Im so lost.

125. anonymous

:( Oh dear.

126. anonymous

Yes.... so what do I need to do to finish using the method you did newton?

127. anonymous

It's OK, I finished it (I only posted the answer because you already had it).

128. anonymous

So, Scotty, are you okay with using Laplace now?

129. anonymous

Yes better. I need to practice a bit more I think :/

130. anonymous

Sure.

131. anonymous

It's a really useful method. You transform a differential equation into an algebraic one which can be easier to solve. Plus, you're initial conditions are taken into account in the process.

132. anonymous

*your

133. anonymous

Damn! Was about to correct that. I'll get you next time...

134. anonymous

Yeah I can see how it would be useful, I just need a little more practice recognizing the transform tables and how to apply them. Also I guess my notation is different than the traditional method

135. anonymous

haha

136. anonymous

You should also practice working them out yourself (i.e. that table entries).

137. anonymous

I wouldn't veen know how to start that, I'll just read some textbooks. Anyways off to the next problem on my review. Thanks guys

138. anonymous

np. I'm off to bed. Good work, Anwar :)

139. anonymous

...and Scotty :)

140. anonymous

Oh thanks :)