Riccati differential equations

Click For Summary

Homework Help Overview

The discussion revolves around finding the general solution to a Riccati differential equation of the form du/dt = u^2 + t^2. The original poster expresses difficulty in approaching this specific form without a known special solution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore substitutions related to Riccati equations, specifically one involving u = y'/Ry, and question the choice of R = -1. There is also a discussion about deriving a second-order linear differential equation from the Riccati form and the validity of the resulting expressions.

Discussion Status

The discussion is active, with participants providing alternative approaches and questioning each other's reasoning. There is acknowledgment of differing interpretations of the derived equations, and suggestions for using power series methods are introduced.

Contextual Notes

Participants note discrepancies in the derived equations and the methods used, indicating a lack of consensus on the correct approach. The original poster's struggle with the absence of a special solution is also highlighted.

mcmaster1987
Messages
2
Reaction score
0
riccati differential equations

--------------------------------------------------------------------------------
how to find general solution of this question

du/dt=u^2+t^2

please say me

i work hard but i do nat know this form of riccati equation. i know when special solution is given however there is no special soltion such that u=u1(t) in this question.
 
Physics news on Phys.org
When I make the usual substitution:

u=\frac{y'}{Ry}

with R=-1 in your case (see info about Riccati equation),

I get:

y''+t^4y=0

Suppose you had to come up with an analytic expression for the eqn. in y. What would you do?
 
jackmell said:
When I make the usual substitution:

u=\frac{y'}{Ry}

with R=-1 in your case (see info about Riccati equation),

I get:

y''+t^4y=0

Suppose you had to come up with an analytic expression for the eqn. in y. What would you do?


why do you make u=y'/Ry and why you take R=-1

please tell me.

and i think, you have made an error process.

because you found an equation y''+yt^4=0

but i found y''+yt^2=0. and then i used second order linear differential equation thecniques.

After that i found y=C1e^it+C2e^(-it). i think this is not true.

Thank you for your interest my question.
 
Ok, my bad. It should be as you said and that's called the parabolic cylinder differential equation:

y''+x^2y=0

But that's not solved using ordinary techniques. You could however, use power series and that's what I was referring to above. Say you get it in the form:

y(x)=\sum_{n=0}^{\infty}a_nx^n

Then the solution to the original DE is:

u(x)=-\frac{\frac{d}{dx} \sum_{n=0}^{\infty}a_nx^n}{\sum_{n=0}^{\infty}a_nx^n}

Nothing wrong with that is there?
 

Similar threads

Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K