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

[Jbreezy]

Homework Statement

dv/dt = (v+1)/10v^2 Euler method delta t = .5, t =0 and v = 1, to find v(4)

Homework 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 homework will be wrong because of calculator mistakes.

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

 

[vela]

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?

 

[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

 

[vela]

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

 

[Jbreezy]

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

 

[vela]

Yes.

 

[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

 

[vela]

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.

 

[Jbreezy]

Step size is 0.2
Sorry I left that out.

 

[vela]

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

 

[Jbreezy]

I get that but for when x = .8

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

 

[Jbreezy]

So my y(1) = 1.233930342

 

[D H]

I get that but for when x = .8

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).

 

[Jbreezy]

Got it thx.