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

[trojansc82]

Homework Statement

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

Homework Equations

Picard's method

The Attempt at a Solution

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

 

[LCKurtz]

Homework Statement

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

Homework Equations

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).

 

[lanedance]

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

 

[lanedance]

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

 

[trojansc82]

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

 

[HallsofIvy]

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.

 

[LCKurtz]

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.

 

[HallsofIvy]

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.

 

[trojansc82]

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

 

[LCKurtz]

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...

 

[trojansc82]

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.

 

[LCKurtz]

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.