# Picard's Method for Solving Differential Equations: Finding y1 = 1 + x

• trojansc82
In summary: What integral? We aren't mind readers. My guess is you have the integral wrong in the first place. Show us your work.
trojansc82

## Homework Statement

y' = -y , y(0) = 1

Picard's method

## The Attempt at a Solution

I found y1 as 1 + 1/2x2, however y1 is really 1 + x

Last edited:
trojansc82 said:

## Homework Statement

y' = -y , y(0) = 1

Picard's method

## The Attempt at a Solution

I found y1 as 1 + 1/2x2, however y1 is really 1 + x

Neither of those is correct for the equation you wrote. Show us what you did so we can find your mistake(s).

you should show you method, so we can see where you're going wrong - what was your first approximation? starting with a constant should integrate to a constant times x and you should have the solution after one iteration

also as LCKurtz points out neither is correct, you should alway check by substituting into the original DE

Sorry I meant y1 is really 1 -x. It was a typo.

I think LCKurtz and lanedance are misunderstanding your notation. Neither $y= 1+ x^2/2$ nor $y= 1+ x$ is a solution to the equation but you are not claiming it is.

Picard's method of solving the differential equation y' = f(x,y), with initial condition $y(0)= y_0$ is an iterative method. Taking $y_0$ to be the initial value, $y_1= \int f(x, y_0)dx$ is the first iteration, then $y_2= \int f(x, y_1(x))dx$, $y_3= \int f(x,y_2(x))dx$, etc.

For the problem as given, the first iteration of Picard's method is, indeed, $y_1(x)= x+ 1$. But no one can show you where you went wrong until you show us what you did.

HallsofIvy said:
I think LCKurtz and lanedance are misunderstanding your notation. Neither $y= 1+ x^2/2$ nor $y= 1+ x$ is a solution to the equation but you are not claiming it is.

Nor are we claiming that.

Picard's method of solving the differential equation y' = f(x,y), with initial condition $y(0)= y_0$ is an iterative method. Taking $y_0$ to be the initial value, $y_1= \int f(x, y_0)dx$ is the first iteration, then $y_2= \int f(x, y_1(x))dx$, $y_3= \int f(x,y_2(x))dx$, etc.

For the problem as given, the first iteration of Picard's method is, indeed, $y_1(x)= x+ 1$.

No, it isn't.

Oh, I completely misread the equation! It is y'= -y, not y'= y as I was seeing!

My apologies to both LCKurtz and lanedance.

trojansc82, I still don't see how you got "1+ x^2/2". Please show us what you did.

HallsofIvy said:
Oh, I completely misread the equation! It is y'= -y, not y'= y as I was seeing!

My apologies to both LCKurtz and lanedance.

trojansc82, I still don't see how you got "1+ x^2/2". Please show us what you did.

Again, I apologize, I mistyped the answer.

The answer for y1 = 1 - x

trojansc82 said:
Again, I apologize, I mistyped the answer.

The answer for y1 = 1 - x

Is that the answer you got? In your original post you said you got something else. Have you figured it out? You never did show us what you did...

LCKurtz said:
Is that the answer you got? In your original post you said you got something else. Have you figured it out? You never did show us what you did...

Within the integral I multiplied -1 (since y was -y) by t. I ended up integrating -t, which came to -1/2 x2.

I have trouble within the integral.

trojansc82 said:
Within the integral I multiplied -1 (since y was -y) by t. I ended up integrating -t, which came to -1/2 x2.

I have trouble within the integral.

What integral? We aren't mind readers. My guess is you have the integral wrong in the first place. Show us your work.

## What is Picard's method for solving differential equations?

Picard's method is an iterative algorithm used to solve initial value problems for ordinary differential equations. It involves breaking down the original problem into a series of smaller problems and solving them step by step.

## How does Picard's method work?

Picard's method involves starting with an initial guess for the solution, then using this guess to approximate the original equation. This approximation is then used to generate a new guess, and the process is repeated until the solution converges to a certain level of accuracy.

## Why is it important to find y1 = 1 + x in Picard's method?

The initial value y1 = 1 + x is used as the starting point for the iterative process in Picard's method. It serves as the first approximation for the solution and is essential for finding subsequent approximations.

## What types of differential equations can be solved using Picard's method?

Picard's method is primarily used for solving first-order ordinary differential equations. It can also be used for certain types of higher-order differential equations by converting them into a system of first-order equations.

## Are there any limitations to using Picard's method for solving differential equations?

While Picard's method is a powerful and widely used technique, it does have its limitations. It may not always converge to a solution, and even when it does, the solution may not be unique. Additionally, it may be computationally intensive for highly nonlinear or complex equations.

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