Question

Can someone please help. I know the answer I just want to know how to do it.

Answer 1

Solve the IVP numerically on the suggested interval, if given, using various step sizes. Compare with values of exact solutions when possible.
y'=3t^(2)-y y(0)=1 [0,1]

Answer 2

this is the answer:
Using step size 0.1 and Euler's method y(1)=1.06121, smaller steps give higher yn(tn)

Answer 3

Let's drop / forget / ignore the answer and focus on how to solve this problem numerically, as well as through techniques of integration.

By y'=3t^(2)-y do you mean\[y' = \frac{ dy }{ dx }=3t^2-y?\]

Answer 4

Have you used Euler's method before? If so, describe very briefly, in your own words, how this method "works."

Answer 5

Excuse me ... I meant\[y' = \frac{ dy }{ dt }=3t^2-y\]...with t as independent variable, y as dependent variable.

Answer 6

yes that is what i mean and yes I barely started learning about Euler's method. The equation is y1=y0+hf(t0+y0). In this case i think y0 is 1 and we plug in 1 for y so we can find the f(t0+y0) since this is the fuction?

Answer 7

close enough. I see you do have the beginnings of understanding of Euler's method.

y(0)=1 [0,1] is redundant. Both say, "If t-0, y=1."

While not strictly necessary, you might want to draw this situation: draw a set of coord. axes and then plot (0,1).

Answer 8

\[\frac{ dy }{ dt }=3t^2-y\] gives you the slope of the tangent line tangent to the curve at any point on the curve. For example, your beginning point is (0,1), meaning that t=0 and y=1. What is the slope of the tangent line to the curve at (0,1)?

Answer 9

1? Im not sure

Answer 10

Replace t by 0 and y by 1 in \[y' = \frac{ dy }{ dt }=3t^2-y\] This results in \[dy/dt = 3(0)^2-1, ~or~?\]

Answer 11

oh is -1

Answer 12

Right. So, picture the point (0,1) and draw a little dash through it in such a way that the run is 1 and the rise -1. It'll slope downward as you move from left to right, right?

Answer 13

yea

Answer 14

You are to increase x in jumps of 0.1. You're to use the fact that dt = delta t =0.1 and the fact that you begin at the point (0,1) to determine the next y value.

I think you've already demonstrated that you have some knowledge of this.

Use the equation\[y _{n+1}=y _{n}+\frac{ dy }{ dt }\Delta t\] to calculate the "new" t value.

Answer 15

At the risk of repeating myself: Your initial t and y values are 0 and 1 respectively.
You've found that the slope of the t. l. at (0,1) is -1.

Delta t is 0.1.

what's the new y-value?

Answer 16

is it y1=1+(0.1)(-1)

Answer 17

Yes, which simplifies to 1-0.1, right? which simplifies to ... ?

Answer 18

0.9

Answer 19

Right. So, your first point was (0,1). Your second point is (0.1,0.9). Agree or disagree?

Answer 20

agree

Answer 21

\[y _{n+1}=y _{n}+\frac{ dy }{ dt }\Delta t\] will now give you your third point. You are now starting with t=0.1, y=0.9, and dy/dt given by which equation?

Answer 22

y2=0.9+(0.1)(-0.9)

Answer 23

Your slope (-0.9) looks reasonable, so i won't insist that you show me where it came from.
Have you any questions about this process?
You have to continue incrementing t until you reach t=1.

Answer 24

when i keep going the y values go down each time

Answer 25

I always do this sort of problem on Excel.
You'd need more points to verify the behavior pattern of the graph. It surely does sound like a decaying exponential function, don't you think?

Answer 26

so what the answer is saying that if we reach t=1 or y(1) it will equal 1.0621?

Answer 27

If you do the arithmetic correctly, you should end up with y=1.0621 when t=1. Caution: I have not done the problem myself; I'm building upon what you have already shared with me.

Answer 28

oh ok and I notice the question says we can use a different step size so does that mean we can choose another number and not use 0.1 all the time?

Answer 29

Supposing that (1,1.0621) is the correct answer when this problem is done numerically (that is, using Euler's method), you are to solve the given D. E. using integration (actually, by the method called "First Order Linear Differential Equations") and initial values; once you finish that, you'll substitute 1 for t and see what value your solution produces for y.

Answer 30

To answer your question: Yes, you can use any reasonable value other than 0.1 for your step size. Note that such value has to be greater than zero; otherwise t would not increase. You could use step size 0.01 and get a considerably more accurate value for y, but this would take 100 or 101 calculations of y, each one dependent on re-calculating the slope, dy/dt.

Answer 31

I need to get off the Internet, but will likely be back on OpenStudy later today.

Answer 32

okay and i plugged in 1 and 1.06121 and got y'=to be 1.93879

Answer 33

You plugged in 1 for which variable (t or y) and got what? you say you 'got y' to be 1.93879. Ask yourself whether that seems right: you are saying that the slope fo the tangent line to the curve when t=1 is +1.93879.

Answer 34

i did not understand what you were trying to say "First Order Linear Differential Equations" do i take the derivative first then plug 1 into t or do I just plug it in like that into the equation

Answer 35

ok

Answer 36

Copy down my address; I'm deleting that post for privacy reasons.

Answer 37

"First order linear differential equations" is a CLASS of differential equations that are relatively easy to solve; another kind is "Separable Differential Equations."
I won't have time to discuss these now, but would be happy to do so later. If you like,
email your question to me at -- , or simply type them right here on OpenStudy and tag me.