# Dv/dt = (v+1)/10v^2 Euler's method

1. Oct 25, 2013

### Jbreezy

1. The problem statement, all variables and given/known data
dv/dt = (v+1)/10v^2 Euler method delta t = .5, t =0 and v = 1, to find v(4)

2. Relevant equations

I'm using v = v(old) + (stepsize)(dv/dt) I'm just wondering if anyone can confirm my results. My teacher said that although this seems easy over half the hw will be wrong because of calculator mistakes.

3. The attempt at a solution

I know it is boring but could someone double check me? Maybe put it in wolfram Idk exactly how to do it.THanks Or maybe someone has a program.

For my final calculation I got v(4) = 1.731618162

Man I hope that is OK

2. Oct 25, 2013

### vela

Staff Emeritus
I got v(4) = 1.5740 using a spreadsheet. Just to make sure, it's $\frac{dv}{dt} = \frac{v+1}{10v^2}$ and v(0)=1, right?

3. Oct 25, 2013

### Jbreezy

The exact question is
Use Euler's method with $\frac{dv}{dt} = \frac{v+1}{10v^2}$ and delta t = 0.5, starting at t =0 and v = 1, to find v(4).
So yeah. But did you use the right step size.I will give you a couple values I want to see if I was doing it right but went wrong.

So for t = 1.0 I have v = 1.18677686
t = 2.0 I have v = 1.406046194
t = 3.0 I have v = 1.524220178
Do any of those look OK?
Thanks

4. Oct 25, 2013

### vela

Staff Emeritus
Your result for t=1.0 matches mine, but the others don't.

5. Oct 25, 2013

### Jbreezy

Figures.
So dv/dt for 1.0 I have .1552624125 is that what you got?

6. Oct 25, 2013

### vela

Staff Emeritus
Yes.

7. Oct 25, 2013

### Jbreezy

Dude I GOT IT! Whohoo. Thanks. It is so funny that something that seems very simple ends up being clustered for me.

If you did it on a spread sheet and it was easy can you punch in another one for me? This is the last one I had from my hw.
If not and it is a drag to do that's OK.

But here it is just in case.

y' = xy -x^2 and y(0) = 1 they want y(1).

I got 3.08106496

8. Oct 25, 2013

### vela

Staff Emeritus
What step size?

With a step size of 0.1, I get about 1.217, and the actual value is about 1.238. It looks like your number is a bit off.

Last edited: Oct 25, 2013
9. Oct 26, 2013

### Jbreezy

Step size is 0.2
Sorry I left that out.

10. Oct 26, 2013

### vela

Staff Emeritus
With a step size of 0.2, I get y(1)=1.195.

11. Oct 26, 2013

### Jbreezy

I get that but for when x = .8

Yours started at 0? Y(0) = 1?

12. Oct 26, 2013

### Jbreezy

So my y(1) = 1.233930342

13. Oct 26, 2013

### D H

Staff Emeritus
Apparently you are updating the independent variable (x) before you are updating the dependent variable (y). Don't do that. The derivative function is f(x,y), not f(x+Δx,y).

14. Oct 26, 2013

### Jbreezy

Got it thx.

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