Riccati Equation. Unsure what substitution to use

  • Thread starter Trestal
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  • #1
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Main Question or Discussion Point

Hi,
I have the equation
y' = y^2 + x^2 and am asked to linearise the equation with the appropriate substitution and then solve the resulting 2nd order linear equation.

My issue is I am unsure what to substitute in for y. I can't seem to find a choice for y which the differential will be a higher order so that I can make the aux equation to solve for some value c.
 

Answers and Replies

  • #2
798
34
Hl!

Let y(x) = -(1/f(x))*(df/dx)
leads to (d²f/dx²)+x²f(x)=0
 
  • #3
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once you have the second order linear ODE, how do you solve it and then get back to the original Riccati equation? I tried power series method but it looks messy and I'm not sure if it's right, I then tried expressing as a Bessel function and get f= c1 x^1/2 J 1/4 (1/2 x^2) + c2 x^1/2 J -1/4 (1/2 x^2), but then how to return to a function in y?
 
  • #4
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Going back to the power series soln, I get a soln for f, then I differentiated f and then substituted it all back into y = -1 f'/f to get a soln for y. The soln for y has power series on numerator and denominator and have different powers of x thru it.
 
  • #5
798
34
Hi !
I agree with your result :
f= c1 x^1/2 J 1/4 (1/2 x^2) + c2 x^1/2 J -1/4 (1/2 x^2)
To return to a function in y, you have to derivate f(x) and bring it back into y = -f'/f
This will leads to a fraction in which the numerator and the denominator are the sum of Bessel fonctions. This is the simplest way to express the result on a closed form.
 

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