y'+f(t)y+y^2 = const it's a riccati equation with only f(t) being non-constant. I know if I have one solution I can find the general solution. But how can I find a particular solution?! isn't these kind of equations standard?!
Hello tucuman87, These equations are not at all that easy to solve if a general function is present. The only way that I know of, for finding a solution is by transforming it into a linear one. This can be done by a transformation of the dependent variable. So, in general the Riccati equation is: [tex]\frac{dy}{dt}+Q(t)y+R(t)y^2=P(t)[/tex] Use now the substitution: [tex]y(t)=\frac{1}{R(t)u(t)}\frac{du(t)}{dt}[/tex] Calculate from this the derivative and put it into the Riccati equation and you will end up with a linear one. Now in your case we have: [tex]y(t)=\frac{1}{u(t)}\frac{du(t)}{dt}[/tex] [tex]\frac{dy(t)}{dt}=-\frac{1}{(u(t))^2}\left(\frac{du(t)}{dt}\right)^2+ \frac{1}{u(t)}\frac{d^2u(t)}{dt^2}[/tex] Giving thus as a transformed equation (without the independent variable notation): [tex]\frac{d^2u}{dt^2}+f(t)\frac{du}{dt}-Cu=0[/tex] As you can see this is an equation with an enormous amount of possible solutions depending on the function f(t). If you have any information on this function, I might be able to help you a bit further. Hope this helps so far.
well... actually... I've checked the "Handbook of Exact Solutions for ODE's" and I saw there that it is possible to to transform this Ricatti equation into a second order linear one. However, my case is not the most general Riccati equation, since the only non-constant function is f(t). I'm looking for an explicit solution, in form of an integral... I know f(t) - it's just an ugly function. The problem is that I cant find a special case like this.. it still falls only in the most general category!
I personally do not think that for a general f(t) you can have a solution by quadrature. Sometime we need the help of special functions.
I have a similar problem with a Riccati equation. The DE I'm trying to solve look quite simple. [tex]\frac{dy}{dt} - y^2= t^2[/tex] Using the given substitution, I obtain [tex]\frac{d^2u}{dt^2}+t^2u=0[/tex] How is this DE related to the Bessel equation? One web site give the following as solution [tex]y=-t\frac{BesselJ(-3/4,t^2/2)C+BesselY(-3/4,t^2/2)}{BesselJ(1/4,t^2/2)C+BesselY(1/4,t^2/2)C} [/tex] OK they used MAPLE. Can we solve it manually? Of course we can if we know the algorithm. I think I can solve [tex]\frac{d^2u}{dt^2}+t^2u=0[/tex] using power series. But how to express the power series in term of special functions ?
If you give the actual function, I might be able to get you one step further because I do not see a solution with the general function f(t). The substitution is applied correctly. The new equation you have now can be solved by comparing it to the following one: [tex]t^2u''+(2k+1)tu'+(\alpha^2t^{2r}+\beta^2)u=0[/tex] Which has the following solution: [tex]u(t)=t^{-k}\left(AJ_{\frac{\gamma}{r}}\left(\frac{\alpha}{r}t^r\right) +BY_{\frac{\gamma}{r}}\left(\frac{\alpha}{r}t^r\right)\right)[/tex] with: [tex]\gamma=\sqrt{k^2-\beta^2}[/tex] For the equation at hand the solution becomes: [tex]u(t)=t^{\frac{1}{2}}\left(AJ_{\frac{1}{4}}\left(\frac{1}{2}t^2\right) +BY_{\frac{1}{4}}\left(\frac{1}{2}t^2\right)\right)[/tex] Using now the inverse transformation, you can obtain the one given in the post. You need to calculate the derivative of the bessel functions and use some properties of them. I haven't done this though. The equation I presented can be obtained by applying a substitution of the independent and dependent variable of the general differential equation of Bessel. This is however an involved algebraic operation. The equation is very general and can be used to find a number of other ones by inspecting the coefficients.
Thanks a lot coomast. I will work on the detail. Can we just replace [tex]BesselY_\frac{1}{4}(t)[/tex] with [tex]BesselJ_\frac{-1}{4}(t)[/tex] since the order of the Bessel function is not an integer ? Dealing with Bessel functions of the second kind is much more difficult.
You can replace the Bessel function of the second kind with one of the first kind, but do keep in mind what you are doing. This because it is not just a replacement. Follow the next steps: [tex]Y_{\frac{1}{4}}\left(X\right)=\frac{J_{\frac{1}{4}}\left(X\right) cos(\frac{\pi}{4})- J_{-\frac{1}{4}}\left(X\right)}{sin(\frac{\pi}{4})}[/tex] This is: [tex]Y_{\frac{1}{4}}\left(X\right)=J_{\frac{1}{4}}\left(X\right) -\sqrt{2} J_{-\frac{1}{4}}\left(X\right)[/tex] Putting this into the solution gives you: [tex]u(t)=t^{\frac{1}{2}}\left((A+B)J_{\frac{1}{4}}\left(\frac{1}{2}t^2\right) -\sqrt{2}BJ_{-\frac{1}{4}}\left(\frac{1}{2}t^2\right)\right)[/tex] Or: [tex]u(t)=t^{\frac{1}{2}}\left(A_0J_{\frac{1}{4}}\left(\frac{1}{2}t^2\right) +B_0J_{-\frac{1}{4}}\left(\frac{1}{2}t^2\right)\right)[/tex] With two new constants of integration. So, it is not replacing one function by the other but a somewhat more involved operation. This is normally done at the end of the solution of the Bessel equation and regarded as trivial, therefore sometimes mistakes can appear if one forget these steps.
So I presume that we can replace [tex]BesselY(1/4,t^2/2)[/tex] with [tex]BesselJ(-1/4,t^2/2)[/tex] when solving [tex]\frac{d^2u}{dt^2}+t^2u=0[/tex] since this is just an intermediate step. But we cannot just simply replace [tex]BesselY(1/4,t^2/2)[/tex] with [tex]BesselJ(-1/4,t^2/2)[/tex] in the final answer [tex]y=-t\frac{BesselJ(-3/4,t^2/2)C+BesselY(-3/4,t^2/2)}{BesselJ(1/4,t^2/2)C+BesselY(1/4,t^2/2)} [/tex] I have cross checked the answer with a different software mathematica. The answer given is (if I copied correctly) [tex]y=\frac{t^2 J(-3/4,t^2/2)+ J(1/4,t^2/2) - t^2 J(5/4,t^2/2) + t^2 J(-5/4,t^2/2)C +J(-1/4,t^2/2)C - t^2 J(3/4,t^2/2)C }{-2t J(1/4,t^2/2) - 2t J(-1/4,t^2/2)C} [/tex] There is no BesselY function.
Hello matematikawan, That answer seems to be the correct one. You might consider to "simplify" it to the first result given, but I wouldn't do it, it's perfectly valid as it is. Coomast
any other methods? Is there any other methods of solving this equation without involving the Bessel equation?
Original problem: Solve [tex]\frac{dy}{dt} = t^2 + y^2 [/tex] Using the substitution given by coomast, we can reduced this Riccati equation to the second order linear DE [tex]\frac{d^2u}{dt^2}+t^2u=0[/tex] Since t=0 is an ordinary point, we should be able to solve this equation using the power series method [tex]u(t) = \Sigma a_n t^n [/tex]. But we won't get a closed form solution.
In general can we relate a Riccati equation [tex]\frac{dy}{dt} + Q(t)y + R(t)y^2 = P(t)[/tex] to a generalized Bessel equation [tex]t^2u''+(2k+1)tu'+(\alpha^2t^{2r}+\beta^2)u=0[/tex] ? What are the functions P(t), Q(t), and R(t) such that the Riccati equation has a solution that can be expressed in term of Bessel functions (or whatever special functions) ? The aim is to determine whether the original problem post by tucuman87 [tex]\frac{dy}{dt} + f(t)y + y^2 = C[/tex] has a special function solution.