Question

Differential Equations Question

Find approximate values of the solution of the given initial value problem at t=0.1,0.2,0.3,0.4 using the Euler method with h=0.1

Find the solution \(y=\phi(t)\) of the given problem and evaluate \(\phi(t)\) at t=0.1,0.2,0.3 and 0.4

\(y\prime = 3\cos(t)-2y\)

Answer 1

\[\Large y' = 3\cos(t) - y\]

Answer 2

Well, let's work on the hard part first, actually solving this this differential equation... or have you already done that?

Answer 3

-2y*

Answer 4

Whoopsie
\[\Large y' = 3\cos(t) - 2y\]
So HAVE you solved this? :D

Answer 5

\(\dfrac{dy}{dt}=3\cos(t)-2y\)
\(dy=(3\cos(t)-2y)dt\)

\(\int dy=\int (3\cos(t)-2y)dt\)

Correct so far?

Answer 6

uhh.. no.... there's a y-term in the right-side... one does not simply integrate that with respect to t...

Answer 7

Well, I am drawing a blank then..... :/

Answer 8

this looks like another one of them differentials that need integrating factors...

Answer 9

\[\Large y' + 2y = 3\cos(t)\]

Answer 10

Integrating factor would be \[\LARGE e^{\int2dt}\]

Answer 11

\(\Large{\mu(t)=e^{2t}}\)

\(\Large{e^{2t}y^{\prime}}+2e^{2t}y=e^{2t}3\cos(t)\)

Look ok so far?

Answer 12

Yup :)

Answer 13

Left side is
\[\Large d(e^{2t}y)= 3e^{2t}\cos(t)dt\]

Answer 14

I'll leave you to do the horrid integration.. :3

Answer 15

Gimme a minute......

Answer 16

<starts stopwatch>

Answer 17

\(\Large{y=\dfrac{3(\cos(t)+\sin(t))}{5}}\)

Answer 18

Alright, so we have solved this lovely little differential equation. What now?

Answer 19

You forgot the constant :3

Answer 20

I am looking at an example, and they aren't showing a constant..... Oh well.

\(\Large{y=\dfrac{3(\cos(t)+\sin(t))}{5}+C}\)

Answer 21

Crud... I meant
\[\Large y e^{2t}= \frac{3\color{red}{e^{2t}}[\cos(t)+ \sin(t)]}{5}+C\]

Answer 22

Wait a sec... rewind... could you redo that integration? Wolfie doesn't agree with you...

Answer 23

\[\Large d(e^{2t}y)= 3e^{2t}\cos(t)dt\]

Answer 24

I made a typo, but it does agree, \(\color{blue}{2\cos(t)}\)*

Answer 25

Okay, so we have this, aye?
\[\Large y e^{2t}= \frac{3\color{}{e^{2t}}[2\cos(t)+ \sin(t)]}{5}+C\]

Answer 26

Now, you had an initial condition?

Answer 27

\(y(0)=0\)
Most interesting initial condition ever.

Answer 28

Oh duh, plug that bad boy in and solve or C yeah?

Answer 29

Yup :D

Answer 30

Don't lament y(0) = 0 as an initial condition... it usually simplifies a lot of things XD

Answer 31

\(\Large{ye^{2t}=3e^{2t}\dfrac{[2cos(t)+sin(t)]}{5}-\dfrac{6}{5}}\)

Answer 32

Okay... so more on those later.. I'll let you worry about checking about individual points t = 0.1 and so on... do we proceed to using Euler's method?

Answer 33

Umm..... honestly? I have no clue. I am looking at the example and it make no sense.

Answer 34

What makes no sense? Euler's Method?

Answer 35

Yeah, this text book and how it is written makes no sense. It kinda jumps around a lot.

Answer 36

Okay, well, it's already well set-up for Euler's method...
Let's look at the original statement of the problem
\[\Large y' = 3\cos(t) - 2y\]

Answer 37

\[\Large y'(t) = f(t,y(t))\] right?

Answer 38

Namely, that the derivative of y with respect to t is a function of t and y. Catch me so far?

Answer 39

I follow me thinks. :)

Answer 40

Okay, first we select a step-size, h.
The smaller, the more accurate.

The choices of t being 0.1, 0.2, 0.3, 0.4 seem to imply that we're to choose a step-size of 0.1. You okay with that, or you want 0.05 for more accuracy?

Answer 41

Or is accuracy not an issue and/or you are specifically instructed to use a step size of h = 0.1?

Answer 42

Find approximate values of the solution of the given initial value problem at t=0.1,0.2,0.3,0.4 using the Euler method with \(\color{blue}{h=0.1}\)
I am guessing that I am supposed to use that.

Answer 43

Okay, well, simpler that way, anyway :)

Answer 44

So, this is the heart and soul of the Euler method:
\[\Large y_{n+1} =y_n + hf(t_n,y_n) \]does that look daunting? :D

Answer 45

Since the step size is h = 0.1, then we have
\[\Large \left.\begin{matrix}t_0 &0\\t_1&0.1\\t_2&0.2\\t_3&0.3\\t_4&0.4\end{matrix}\right.\]

Answer 46

So, ready to begin?

Answer 47

@austinL don't worry, it's a lot easier than it makes itself out to be... so... do we proceed? :D

Answer 48

Yes, sorry. I went and made some coffee! :D Yep, good to go.

Answer 49

Okay, we want to know the approximation for \(\large y_1\)
so, we plug in :)
\[\Large y_{n+1} =y_n + hf(t_n,y_n)\]
\[\Large y_{1} =y_0 + hf(t_0,y_0)\]Got it?

Answer 50

Yep. :)

Answer 51

This is pretty much the only time where you actually *need* that initial condition.
But it's pretty darn important, as without it, you don't have a place to start XD

So you know that y(0) = 0
that means
\[\Large y_0 =0\]\[\Large t_0=0\]
Now solve for \(\large y_1\)

Answer 52

You also might want to have a calculator handy

Answer 53

\(y_1=y_0+hf(t_0,y_0)=0+(0.1)(3cos(0)-2(0))=0.1\)
?

Answer 54

As I recall... 3cos(0) = 3...

Answer 55

Oh duh... silly error. 0.3

Answer 56

2*0=0 Austin, get it together.

Answer 57

Okay, great :D
Now you have \(\large y_1\) which is an approximation for \(\large y(t_1)=y(0.1)\)
Great:D

Answer 58

Can you predict the next step? Work on \(\large y_2\)
\[\Large y_{2} =y_1 + hf(t_1,y_1)\]

Answer 59

Note that it's y1 and t1 now, since you now know the value of y1.

Answer 60

Would it be 0.238501?

Answer 61

I.... hang on :D

Answer 62

How did you do this?

Answer 63

\(t_1=0.1,~~y_1=0.3,~~h=0.1\)
Plug it in?

Answer 64

I get 0.658 something

Answer 65

\(\Large{y_2=0.3+(0.1)(3\cos(0.1)-2(0.3))}\)
I just got a different answer, but is this the correct format?

Answer 66

Should be.

Answer 67

hang on again...

Answer 68

I definitely get 0.6585 something
try it again.

Answer 69

I get 0.538501
What am I doing wrong D:

Answer 70

google
https://www.google.com/#q=0.3+%2B+(0.1)(3cos(0.1)+%2B+2(0.3))

Answer 71

Ok, what the deuce...

Answer 72

I really don't know...

Answer 73

that is... why you're not getting it... why don't you list down your steps here, one at a time

Answer 74

For the next term I have, 1.08423

Answer 75

Sorry, what?
You used y1 = 0.658something?

Answer 76

And yes, that's right :)

Answer 77

Yep, y2=0.658501

Answer 78

So you finally got it then? :D

Answer 79

I do believe so! Thanks very much!

Answer 80

You have y1, y2, y3 all down.
finish it off with y4

Answer 81

@austinL hey don't leave me hanging, buddy D:

Answer 82

I am just finishing up.
\(t=~~~~~~~h=0.1\)
\(0.0~~~~~~~~~0.3\\
0.1~~~~~~~~~0.658501\\
0.2~~~~~~~~~1.08423\\
0.3~~~~~~~~~1.58768\\
0.4~~~~~~~~~2.18153\)

Answer 83

a bit off...

Answer 84

2.18153... is actually y5 (or when t = 0.5)

Answer 85

Don't need t=0.5

Answer 86

Earth to austin... when t = 0, y should be also be 0
Like... you know, your initial value? XD

Answer 87

Adjusting...
\[0.1~~~~~~~~~0.3\\
0.2~~~~~~~~~0.658501\\
0.3~~~~~~~~~1.08423\\
0.4~~~~~~~~~1.58768\\
0.4~~~~~~~~~2.18153\]

Answer 88

omfg.... *facepalm*

Answer 89

Good thing I asked for a confirm. But you've pretty much gotten the Euler method down again, right? :D

Answer 90

Yeah, I feel much better about it :)

Answer 91

Whoops...
\[0.1~~~~~~~~~0.3\\
0.2~~~~~~~~~0.658501\\
0.3~~~~~~~~~1.08423\\
0.4~~~~~~~~~1.58768\\
\color{red}{0.5}~~~~~~~~~2.18153\\TJ ~~~~~~~~~awsum \ \text{^_^ }\]

Answer 92

Thank you very very very much!!!

Answer 93

No problem... now time for me to catch some sleep :)
Nighty night ^_^
-----------------------------
Terence out

Answer 94

Nighty night!