Question

I am having trouble understanding Riccati's EQ for first order ODE. Here is the ex in my book:

y'+y^2=t^2-2t

I don't understand how the sub will work.

y'+(at+b)^2=t^2-2t
a^2t^2+2abt+b^2

Equating Coefficients:
a^2=1
2ab=-2

What happens w/ b^2?

Answer 1

Assume \(y=at+b\implies y'=a\). Substitute into our Ricatti equation:$$a+(at+b)^2=t^2-2t\\a+a^2t^2+2abt+b^2=t^2-2t$$Now we equate corresponding coefficients for the powers of \(t\), i.e. \(t^0\) (for constants), \(t\), \(t^2\).
$$\color{red}{a^2}t^2+\color{blue}{2ab}t+\color{green} {(a+b^2)}=\color{red}1t^2+\color{blue}{-2}t+\color{green}{0}\quad\implies\quad a^2=1,2ab=-2,a+b^2=0$$

Answer 2

y' is sub for a in every Ricatti?

Answer 3

How do you know a= 1 and not -1?

Answer 4

\(a\) has to be \(-1\): \(a+b^2=0\implies a=-b^2\)
Since \(b^2\) is positive for all real \(b\), \(a\) must be negative. From \(a^2=1\) we conclude \(a=-1\) and then from \(2ab=-2\) we conclude \(b=1\).

Answer 5

\(y'=a\) only because we assumed \(y=at+b\) to get a particular solution to our equation (using the method of undetermined coefficients).

Answer 6

Okay- I understand

Answer 7

also I don't really have a consistent schedule unfortunately but if I see more posts I'll be happy to help you!

Answer 8

Okay- thanks :)