Calculate the first four Picard Iterates of the equation y' - y = x^2

[brit123]

Help Please!

Calculate the first four Picard Iterates of the equation y' - y = x^2 with the condition y(0) = -1

and it was given that y'(x) = x^2 +y and y(0)= -1



Need a little help with this question... Not sure what to do.

 

[AiRAVATA]

The definition of Picard iterates is

$$\phi_0=y(x_0),$$

$$\phi_{n+1}(x)=\phi_0+\int_{x_0}^x f(s,\phi_n(s))ds,$$

where $f$ is given by

[tex]y'(x)=f(x,y(x)).[/itex]

In your case, $f(x,y(x))=x^2+y$, $\phi_0=y(0)=-1$, and

$$\phi_1=-1+\int_0^x (s^2+\phi_0)ds=-1+\int_0^x (s^2-1)ds.$$

What is $\phi_2$? and $\phi_n$?

 

[tacman]

To solve this problem, we can use the Picard Iteration method, which is an iterative technique used to approximate a solution to a first-order differential equation. The first four Picard Iterates are given by the following formulas:

y_0(x) = y(0) = -1

y_1(x) = y(0) + ∫x^2 dx = -1 + (x^3)/3

y_2(x) = y(0) + ∫(x^2 + y_1(x)) dx = -1 + (x^3)/3 + (x^5)/5 + (y_1(x))^2/2

y_3(x) = y(0) + ∫(x^2 + y_2(x)) dx = -1 + (x^3)/3 + (x^5)/5 + (x^7)/7 + (y_2(x))^2/2

To find the first four Picard Iterates, we need to substitute the given condition y(0) = -1 into each formula. This gives us:

y_0(x) = -1

y_1(x) = -1 + (x^3)/3

y_2(x) = -1 + (x^3)/3 + (x^5)/5 + (-1 + (x^3)/3)^2/2

y_3(x) = -1 + (x^3)/3 + (x^5)/5 + (x^7)/7 + (-1 + (x^3)/3 + (x^5)/5 + (-1 + (x^3)/3)^2/2)^2/2

Simplifying each of these expressions, we get:

y_0(x) = -1

y_1(x) = -1 + (x^3)/3

y_2(x) = -1 + (x^3)/3 + (x^5)/5 + (x^6)/6

y_3(x) = -1 + (x^3)/3 + (x^5)/5 + (x^7)/7 + (x^9)/9 + (x^10)/10 + (x^11)/66

Therefore, the first four Picard Iterates of the equation y' - y = x^2 are:

y_