- #1

yukcream

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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?

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- Thread starter yukcream
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- #1

yukcream

- 59

- 0

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

- #2

- #3

yukcream

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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~

- #4

AKG

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[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 y

[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 y

- #5

yukcream

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AKG said:

[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 y_{1}= 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 y_{1}, and can even express y' in terms of z and y_{1}. 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 can't solve this D.E?

- #6

AKG

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

- #7

saltydog

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AKG said:

"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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