Solving a first order ODE using the Adomian Decomposition method

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The discussion focuses on solving the first-order ordinary differential equation (ODE) y' + y² = -2 with the initial condition y(0) = 0 using the Adomian Decomposition method. Participants share their attempts at finding the Adomian polynomials, with initial calculations yielding y₀ = -2t and y₁ = -4t³/3. There is some confusion regarding notation, particularly between y₀ and y(0), but overall, the steps taken appear correct. The goal is to compute up to five terms of the polynomials for comparison with analytical and numerical methods. The Adomian method is acknowledged as a viable approach for solving both linear and nonlinear ODEs.
chwala
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Homework Statement


how do we solve the ode ## y'+y^2=-2, y(0)=0## using adomian decomposition method?

Homework Equations

The Attempt at a Solution


##Ly = -2-y^2##
## y= 0 + L^{-1}[-2-y^2]##
##y_{0}= -2t##
##y_{1}= -L^{-1}[4t^2] = -4t^3/3## are my steps correct so far in trying to get the Adomian Polynomials? am i following the correct steps?
 
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Oh, not a Pindaric ode then...
 
chwala said:

Homework Statement


how do we solve the ode ## y'+y^2=-2, y(0)=0## using adomian decomposition method?

Homework Equations

The Attempt at a Solution


##Ly = -2-y^2##
## y= 0 + L^-1[-2-y^2]##
##y[0]= -2t##
##y[1]= -L^-{1}[4t^2] = -4t^3/3## are my steps correct so far in trying to get the Adomian Polynomials? am i following the correct steps?
I'm not used to this method, but your example is pretty close to the one on Wikipedia:
https://en.wikipedia.org/wiki/Adomian_decomposition_method
 
I've never heard of the Adomian method, either, but I'm very familiar with using Laplace transforms and inverses to solve diff. equations.
Comments below.
chwala said:

The Attempt at a Solution


##Ly = -2-y^2##
## y= 0 + L^{-1}[-2-y^2]##
##y[0]= -2t##
The notation used in the wiki article whose link fresh_42 provided is ##y_0##. Your notation confused me into thinking you meant y(0), which is given as 0.
chwala said:
##y[1]= -L^{-1}[4t^2] = -4t^3/3## are my steps correct so far in trying to get the Adomian Polynomials? am i following the correct steps?
It looks good so far. How many terms in the polynomial do you need to get?
 
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Mark44 said:
I've never heard of the Adomian method, either, but I'm very familiar with using Laplace transforms and inverses to solve diff. equations.
Comments below.
The notation used in the wiki article whose link fresh_42 provided is ##y_0##. Your notation confused me into thinking you meant y(0), which is given as 0.
It looks good so far. How many terms in the polynomial do you need to get?
i want to try and get up to 5 terms of the polynomials and then probably compare with the analytical method/numerical methods...let me try and get the other polynomials then you can correct where necessary...
 
Mark44 said:
I've never heard of the Adomian method, either, but I'm very familiar with using Laplace transforms and inverses to solve diff. equations.
Comments below.
The notation used in the wiki article whose link fresh_42 provided is ##y_0##. Your notation confused me into thinking you meant y(0), which is given as 0.
It looks good so far. How many terms in the polynomial do you need to get?
in life we are all learner's probably its good to admit that we may not know everything in math or rather i may know something which you do not know...now you know Adomian decomposition method has been used in solving first order and second order linear and non linear ode and pde. I am conversant with the laplace transforms...
 
i am getting
##y_{3}=-16t^5/15##
##y_{4}= -4t^7/5##
is this correct?
 
Last edited:
Mark44 said:
I've never heard of the Adomian method, either, but I'm very familiar with using Laplace transforms and inverses to solve diff. equations.
Comments below.
The notation used in the wiki article whose link fresh_42 provided is ##y_0##. Your notation confused me into thinking you meant y(0), which is given as 0.
It looks good so far. How many terms in the polynomial do you need to get?
i just made some changes on the notation, sorry...
 

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