Riccati differential equations

mcmaster1987
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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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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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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