Question

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

Answer 1

I could use a LaPlace transform right?

Answer 2

Yeah.. and it's pretty easy.

Answer 3

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

Answer 4

Yeah.

Answer 5

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

Answer 6

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

Answer 7

Oh.. duh.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Would it just be y/s ?

Answer 12

or no, I could solve for L(y)

Answer 13

I think I have it

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

are you sure about the simplification you made?

Answer 18

oh I see it.. it's right.

Answer 19

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

Answer 20

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

Answer 21

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:
\[{9+s \over s(s+5)}\]

Answer 22

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

Answer 23

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

Answer 24

not good at ALL with partial fractions...

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

don't*

Answer 29

A(s+5)+B(s)

Answer 30

err

Answer 31

that should equal to?

Answer 32

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

Answer 33

nearly

Answer 34

you multiply both sides by s(s+5)

Answer 35

Oh right ok so its

Answer 36

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

Answer 37

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

Answer 38

you're doing fine :D

Answer 39

I'm 'half' watching...

Answer 40

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

Answer 41

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

Answer 42

you're on the right track

Answer 43

A=9/5

Answer 44

Then I put that back into the original for B?

Answer 45

or A and B for s...

Answer 46

Yeah.. or just plug s=-5

Answer 47

b=-4/5

Answer 48

now what does Y(s) equal to?

Answer 49

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?

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

Isn't it 9+s

Answer 58

oh yeah 9+s sorry

Answer 59

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

Answer 60

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

Answer 61

You do nothing but the inverse Laplace.

Answer 62

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

Answer 63

Yeah.

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

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

Answer 68

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

Answer 69

Am I complicating things here? :(

Answer 70

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

Answer 71

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

Answer 72

yeah.

Answer 73

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

Answer 74

I just think you're not familiar with notation.

Answer 75

e^-at is the transform...

Answer 76

e^-5t then right?

Answer 77

You know what you need to do, though.

Answer 78

Yes

Answer 79

Me?

Answer 80

No, Scotty.

Answer 81

Exactly Scotty.

Answer 82

Sorry...

Answer 83

It's OK :)

Answer 84

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

Answer 85

Which basically will end up e^5t

Answer 86

How?

Answer 87

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

Answer 88

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

Answer 89

It's in x not in t.

Answer 90

Yes, Scotty.

Answer 91

But in x.

Answer 92

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

Answer 93

so e^-(5x)

Answer 94

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

Answer 95

We all are used to t.

Answer 96

perfect

Answer 97

Ok I was rushing through on the simplification.

Answer 98

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

\[\implies c = \frac{1}{5} \ln 4 \implies x = \ln \frac{4}{9-5y} \implies y = \frac{9-4e^{-5x}}{5}\]