Riccati Method

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I want to read a bit more example on using the Riccati Method when solvng D.E, who can help me?
 

AKG

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AKG said:
Thanks very much~
But how can I get one of the soultion of
y'= [2cos^2(x) - sin^2(x) + y^2]/ 2cos(x) =0 ? (the second example in the reading)
If can't~ no use of the Riccati method~
 

AKG

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What's the problem? For one, they give the solution right on the page, so I don't know why you think you can't do it (unless you mean that you're trying to solve it yourself and haven't looked at the answer). Second of all:

[tex]\frac{2\cos ^2(x) - \sin ^2(x) + y^2}{2\cos (x)} = \left (\frac{2\cos ^2(x) - \sin ^2(x)}{2\cos (x)}\right )y^0 + (0)y^1 + \frac{1}{2\cos (x)}y^2[/tex]

Also, I don't know why you're setting it to 0, you don't need to solve y' = 0. You have an equation for y' in terms of f and x, and you make a substitution for z to get you a linear equation which you can solve. You're given that y1 = sin(x) is a solution, so set:

[tex]z = \frac{1}{y - y_1}[/tex]

If you isolate y in that equation, then you can express y in terms of z and y1, and can even express y' in terms of z and y1. You find those expressions and substitute them into your Ricatti equation. You then perform the algebraic manipulations to isolate z', and on the right side you should end up with a linear expression. Solve for z easily, then substitute back to find y.
 
59
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AKG said:
What's the problem? For one, they give the solution right on the page, so I don't know why you think you can't do it (unless you mean that you're trying to solve it yourself and haven't looked at the answer). Second of all:

[tex]\frac{2\cos ^2(x) - \sin ^2(x) + y^2}{2\cos (x)} = \left (\frac{2\cos ^2(x) - \sin ^2(x)}{2\cos (x)}\right )y^0 + (0)y^1 + \frac{1}{2\cos (x)}y^2[/tex]

Also, I don't know why you're setting it to 0, you don't need to solve y' = 0. You have an equation for y' in terms of f and x, and you make a substitution for z to get you a linear equation which you can solve. You're given that y1 = sin(x) is a solution, so set:

[tex]z = \frac{1}{y - y_1}[/tex]

If you isolate y in that equation, then you can express y in terms of z and y1, and can even express y' in terms of z and y1. You find those expressions and substitute them into your Ricatti equation. You then perform the algebraic manipulations to isolate z', and on the right side you should end up with a linear expression. Solve for z easily, then substitute back to find y.
What you means is unless one of the soultion is given i.e y1 otherwise we cant solve this D.E?
 

AKG

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I don't know if you even read the link I gave you, since if you did there should be no question as to how to solve it. You also missed the seventh sentence on that page:

"Without knowing at least one solution, there is absolutely no chance to find any solutions to such an equation."
 

saltydog

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AKG said:
I don't know if you even read the link I gave you, since if you did there should be no question as to how to solve it. You also missed the seventh sentence on that page:

"Without knowing at least one solution, there is absolutely no chance to find any solutions to such an equation."
You know AKG, I really think that statement should be qualified: It depends of course on what P(x), Q(x), and R(x) are. By use of the transformation:

[tex]y(x)=\frac{u^{'}}{Ru}[/tex]

The Ricccati equation is converted to a linear second-order ODE:

[tex]R\frac{d^2u}{dx^2}-(R^{'}-QR)\frac{du}{dx}-PR^2u=0[/tex]

In some cases, this equation can be solved directly or via power series.
 

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