Question

how to find the general solution for this Riccati equation? \[y' = y^2-x^2+1\]
I think I need to find 1 particular solution and after that it's easy to find the general solution, but I don't know how to find the particular one, some help would be nice

Answer 1

let z = y^-2 might be a reasonable technique

Answer 2

er, z = y^2

Answer 3

\[z=y^2~:~z' = 2y y'\]
\[\frac{1}{2\sqrt z}z'-z=1-x^2\]
....then again

Answer 4

i like to run a power series to see meself:

y=\(\sum_0~c_n~x^n\)

y' = \(\sum_1~c_n~nx^{n-1}\)

Answer 5

but that ^2 on the y might be tricky ....

might help if i knew what a riccati eq was

http://www.sosmath.com/diffeq/first/riccati/riccati.html

that looks useful, so i was on the right track of a substitution

Answer 6

hi, that's an interesting way to solve the problem , but I still can't figure out how to continue solving the problem , or how to use power series to solve it, I hope you can help further on this problem , thank you

Ps: I thought I have to find a particular solution in some way to be able to use the method presented on the site you just linked

Answer 7

\(y=x\) is a solution. You can use that to find the particular one.
\[y'=y^2-x^2+1\]
Letting \(y=x\), you have \(y'=1\). Substituting, you have
\[1=x^2-x^2+1,\]
which is true.

Answer 8

I kinda guessed that solution too , the thing is my teacher doesn't really take guessed solutions at exams because "you probably copied it from someone else" . Is there any way/method/algorithm to find that solution for this problem and for similar problems? Thanks for the help ;)

Answer 9

That's pretty unreasonable of your teacher; I don't see how showing why \(y=x\) is a solution, like I did in my last post, shouldn't be accepted.

Answer 10

Also, it's not really a guess. More of an observation.