Picard's method solution vs exact solution

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In summary, the conversation discusses using Picard's method to find an approximate solution to the differential equation dy/dx = 2xy and comparing it to the exact solution. The attempt at a solution involves using Picard's method and solving the DE exactly using separation and integration, with the final answer being e^(x^2) + C. After correcting a mistake, it is determined that the approximate and exact solutions are equal.
  • #1
phosgene
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Homework Statement



I have to use Picard's method to find an approximate solution to

[itex]\frac{dy}{dx}=2xy[/itex]

and then compare it to the exact solution, given that [itex]y(0)=2[/itex]

Homework Equations



Picard's method:

[itex]y_{n}=y_{0} + ∫^{x}_{x_{0}} f(\phi,y_{n-1}(\phi))d\phi[/itex]

The Attempt at a Solution



So I go over Picard's method about 3 times and get the result:

[itex]2(1+x^2 + \frac{x^4}{2} + \frac{x^6}{6})[/itex]

which looks like it miiiight be going towards:

[itex]2(\sum ^{\infty}_{n=0} \frac{x^{2n}}{n!}) = 2e^{x^2}[/itex]I then solve the DE exactly by noting that the DE is separable, so I separate and integrate to obtain:

[itex]y= e^{x^2}+C[/itex]

Using my initial condition, I find that C=1. So, my question is, have I done this correctly? I have a feeling that I'm supposed to get the same answer for both.
 
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  • #2
phosgene said:

Homework Statement



I have to use Picard's method to find an approximate solution to

[itex]\frac{dy}{dx}=2xy[/itex]

and then compare it to the exact solution, given that [itex]y(0)=2[/itex]

Homework Equations



Picard's method:

[itex]y_{n}=y_{0} + ∫^{x}_{x_{0}} f(\phi,y_{n-1}(\phi))d\phi[/itex]

The Attempt at a Solution



So I go over Picard's method about 3 times and get the result:

[itex]2(1+x^2 + \frac{x^4}{2} + \frac{x^6}{6})[/itex]

which looks like it miiiight be going towards:

[itex]2(\sum ^{\infty}_{n=0} \frac{x^{2n}}{n!}) = 2e^{x^2}[/itex]


I then solve the DE exactly by noting that the DE is separable, so I separate and integrate to obtain:

[itex]y= e^{x^2}+C[/itex]

Using my initial condition, I find that C=1. So, my question is, have I done this correctly? I have a feeling that I'm supposed to get the same answer for both.

e^(x^2)+1 is NOT a solution to the ODE. Try it. You messed up where the constant C belongs.
 
  • #3
Argh! Silly mistake. Corrected it and now they are equal, thanks :)
 

1. What is Picard's method solution?

Picard's method is a numerical method used to approximate the solution of a differential equation. It involves breaking down the problem into smaller, simpler equations and solving them iteratively until a satisfactory solution is reached.

2. How does Picard's method solution compare to the exact solution?

Picard's method solution is an approximation of the exact solution and may not be the exact solution itself. It is often used when the exact solution is difficult or impossible to obtain.

3. When is Picard's method solution preferred over an exact solution?

Picard's method solution is preferred when the exact solution is not readily available or when it is too time-consuming or complex to obtain. It is also useful for solving differential equations that do not have closed-form solutions.

4. What are the limitations of Picard's method solution?

Picard's method solution may not always converge to the exact solution and may require a large number of iterations to reach a satisfactory solution. It is also limited to solving first-order differential equations and may not be suitable for higher-order equations.

5. Are there any applications of Picard's method solution in real-world problems?

Yes, Picard's method solution is commonly used in fields such as physics, engineering, and economics to approximate solutions for differential equations that arise in real-world problems. It is also used in computer simulations and numerical analysis.

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