## 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!!

ehild
Homework Helper

## 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

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 hw problems has step size 0.01...

D H
Staff Emeritus

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)?

ehild
Homework Helper
is there any way of solving y(1) without doing it nine more times? because my next hw 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

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?

ehild
Homework Helper
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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