riccati differential equations


by mcmaster1987
Tags: differential, equations, riccati
mcmaster1987
mcmaster1987 is offline
#1
Oct14-10, 11:39 AM
P: 2
riccati differential equations

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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.
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jackmell
jackmell is offline
#2
Oct14-10, 06:23 PM
P: 1,666
When I make the usual substitution:

[tex]u=\frac{y'}{Ry}[/tex]

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

I get:

[tex]y''+t^4y=0[/tex]

Suppose you had to come up with an analytic expression for the eqn. in y. What would you do?
mcmaster1987
mcmaster1987 is offline
#3
Oct15-10, 11:03 AM
P: 2
Quote Quote by jackmell View Post
When I make the usual substitution:

[tex]u=\frac{y'}{Ry}[/tex]

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

I get:

[tex]y''+t^4y=0[/tex]

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.

jackmell
jackmell is offline
#4
Oct15-10, 02:15 PM
P: 1,666

riccati differential equations


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

[tex]y''+x^2y=0[/tex]

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:

[tex]y(x)=\sum_{n=0}^{\infty}a_nx^n[/tex]

Then the solution to the original DE is:

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

Nothing wrong with that is there?


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