anonymous
  • anonymous
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?
Differential Equations
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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), \(t\), \(t^2\). $$\color{red}{a^2}t^2+\color{blue}{2ab}t+\color{green} {(a+b^2)}=\color{red}1t^2+\color{blue}{-2}t+\color{green}{0}\quad\implies\quad a^2=1,2ab=-2,a+b^2=0$$
anonymous
  • anonymous
y' is sub for a in every Ricatti?
anonymous
  • anonymous
How do you know a= 1 and not -1?

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anonymous
  • anonymous
\(a\) 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 \(b=1\).
anonymous
  • anonymous
\(y'=a\) only because we assumed \(y=at+b\) to get a particular solution to our equation (using the method of undetermined coefficients).
anonymous
  • anonymous
Okay- I understand
anonymous
  • anonymous
also I don't really have a consistent schedule unfortunately but if I see more posts I'll be happy to help you!
anonymous
  • anonymous
Okay- thanks :)

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