Simple differential eq'n problem, check my answer please

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Homework Help Overview

The discussion revolves around solving a simple first-order differential equation, dy/dt = 2y + 3, with an initial condition of y(0) = 0.06. Participants are attempting to find the value of y(1) using numerical methods and comparing their results with those obtained from a spreadsheet.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the iterative method for approximating the solution and express concerns about the accuracy of their results. Some question whether it is necessary to perform multiple iterations to reach y(1) and explore the possibility of finding a quicker method. Others mention the use of exact solutions and compare their findings with numerical approximations.

Discussion Status

The conversation is ongoing, with participants sharing their calculations and questioning the validity of their results. Some have suggested that the iterative method may not yield accurate results without a sufficiently small step size, while others are exploring exact solutions. There is no explicit consensus on the best approach yet.

Contextual Notes

Participants note the challenge of achieving accurate results with larger step sizes and express concern over the significant discrepancies between their manual calculations and spreadsheet outputs. The discussion also highlights the potential for error margins in numerical methods.

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Homework Statement



dy/dt=2y+3, y(0)=0.06, find y(1)= ?, Δt=0.1

Homework Equations



y_k+1 = y_k + Δt*f(t_k, y_k)

The Attempt at a Solution



y_1 = y_0 + Δt*f(t_0, y_0)
y_1 = 0.06 + 0.1*(2(0.06)+3)=0.372

y(1) = 0.372??
this doesn't seem right because I did this on excel and i got around a y(1)~4.00. the margin of error seems way to big. help!
 
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nchin said:

Homework Statement



dy/dt=2y+3, y(0)=0.06, find y(1)= ?, Δt=0.1

Homework Equations



y_k+1 = y_k + Δt*f(t_k, y_k)

The Attempt at a Solution



y_1 = y_0 + Δt*f(t_0, y_0)
y_1 = 0.06 + 0.1*(2(0.06)+3)=0.372

y(1) = 0.372??
this doesn't seem right because I did this on excel and i got around a y(1)~4.00. the margin of error seems way to big. help!

You did the first step of iteration only, and calculated y(0.1). Repeat it nine more times. (But it will be a very poor approximation of y(1).)ehild
 
ehild said:
You did the first step of iteration only, and calculated y(0.1). Repeat it nine more times. (But it will be a very poor approximation of y(1).)


ehild


is there any way of solving y(1) without doing it nine more times? because my next homework problems has step size 0.01...
 
Use a spreadsheet.
 
my teacher wants me to compare my approx value with my the spreadsheet value. I was wondering if there was a fast shortcut to calculate by hand y(1) with step sizes like 0.1 or .01? or is the only way to do it is one by one until i reach value y(1)?
 
nchin said:
is there any way of solving y(1) without doing it nine more times? because my next homework problems has step size 0.01...

It is easy to solve the differential equation by an exact way. If you consider y as the independent variable, dt/dy = 1/(2y+3). If the derivative of a function f(y) is 1/(2y+3), what is the function?

t=(1/2) ln|2y+3| + C, y= Ae2t -1.5, with the initial condition, y=1.56e2t -1.5

ehild
 
ehild said:
It is easy to solve the differential equation by an exact way. If you consider y as the independent variable, dt/dy = 1/(2y+3). If the derivative of a function f(y) is 1/(2y+3), what is the function?

t=(1/2) ln|2y+3| + C, y= Ae2t -1.5, with the initial condition, y=1.56e2t -1.5

ehild

oooh i actually did do that but I thought I was wrong because my approx value was y(1)=10.0269, which is way different from the excel value of ~4.00. are the error margins usually this big?
 
what did you make that A variable equal to?
 
nchin said:
oooh i actually did do that but I thought I was wrong because my approx value was y(1)=10.0269, which is way different from the excel value of ~4.00. are the error margins usually this big?

I got the same y(1) ≈10. The simple iterative method you applied is quite poor, you have to use very small Δt-s to get near to the real solution. With Δt=0.01 it would be much better. But there are more advanced methods for numerical solution of differential equations, Runge-Kutta method, for example. All of them involve lot of computations, so you need a program to do them.

ehild
 
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