Solving Differential Equations with Riccati Method

[yukcream]

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

 

[AKG]

Look here.

 

[yukcream]

Look here.

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]

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:

\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

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:

z = \frac{1}{y - y_1}

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.

 

[yukcream]

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:

\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

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:

z = \frac{1}{y - y_1}

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 can't solve this D.E?

 

[AKG]

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]

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:

y(x)=\frac{u^{'}}{Ru}

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

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

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