# General 2nd order ODE with non-constant coefficient

I'm trying to solve the general case of second order ODE: $$y''(t) + p(t)y'(t) + q(t)y(t) = 0$$.

where $$p(t)$$ and $$q(t)$$ are polynomials divides contains some symbolic constant and $$t$$,$$t^2$$,$$t^3$$ and $$t^4$$ elements.

I tried to use MATLAB dsolve but it cannot solve it. is there a recommended book where I can find a way to solve this problem? or a different program like Maple or Mathematica to solve it with a symbolic toolbox?

Thanks.

• Do you want to solve it analytically or numerically? – Arastas Jun 11 at 22:00
• Im looking for analytical solution. for example: y''+(t^2+t+1)/(t^3+2)y'+(t^4)/(t+2)y=0 – Iftahnaf Jun 12 at 5:01

$$y''(t)+p(t)y'(t)+q(t)y(t)=0$$ Analytical solving of this general linear second order ODE is a much too wide question, even if $$p(t)$$ and $$q(t)$$ are not any kind of functions, but polynomial fractions.

For a general approach see : http://mathworld.wolfram.com/Second-OrderOrdinaryDifferentialEquation.html

If you don't want a closed form solution, you can try to find a solution on the form of infinite series.

Often the closed form solution requires some special functions which where defined and standardised especially to solve a particular kind of ODE.

For example in case of $$p(t)=\frac{1}{t}$$ and $$q(t)=\frac{t^2-n^2}{t^2}$$ the analytic solution is $$y(t)=c_1J_n(t)+c_2Y_n(t)$$ $$J_n(t)$$ and $$Y_n(t)$$ are the Bessel functions of first and second kind respectively.

More complicated example : Case of $$p(t)=\frac{c-(a+b+1)t}{t^2-t}$$ and $$q(t)=\frac{ab}{t(t-1)}$$ the analytic solution is $$y(t)=c_1\:\:_2F_1(a,b;c;t)+c_2\:(-t)^{1-c}\:_2F_1(a-c+1,b-c+1;2-c;t)$$ $$\:_2F_1(a,b;c;t)$$ is the Gauss hypergeometric function.

They are a lot of examples of such ODEs which solutions are expressed with convenient special functions. But in the general case of any $$p(t)$$ and $$q(t)$$ the convenient special functions where not always standardized.

In case of your ODE with $$p(t)=\frac{t^2+t^+1}{t^3+2}$$ and $$q(t)=\frac{t^4}{t+2}$$ as far as I know no convenient special function is available. May be a generalized hypergeometric function ? Sorry I have not enough available time to check it and I doubt the extra effort is worth it. As usual in such a situation, one commonly use numerical method for solving.

Hint:

For $$\dfrac{d^2y}{dt^2}+\dfrac{t^2+t+1}{t^3+2}\dfrac{dy}{dt}+\dfrac{t^4}{t+2}y=0$$

Let $$r=\dfrac{1}{t}$$ ,

Then $$\dfrac{dy}{dt}=\dfrac{dy}{dr}\dfrac{dr}{dt}=-\dfrac{1}{t^2}\dfrac{dy}{dr}=-r^2\dfrac{dy}{dr}$$

$$\dfrac{d^2y}{dt^2}=\dfrac{d}{dt}\left(-r^2\dfrac{dy}{dr}\right)=\dfrac{d}{dr}\left(-r^2\dfrac{dy}{dr}\right)\dfrac{dr}{dt}=\left(-r^2\dfrac{d^2y}{dr^2}-2r\dfrac{dy}{dr}\right)(-r^2)=r^4\dfrac{d^2y}{dr^2}+2r^3\dfrac{dy}{dr}$$

$$\therefore r^4\dfrac{d^2y}{dr^2}+2r^3\dfrac{dy}{dr}+\dfrac{\dfrac{1}{r^2}+\dfrac{1}{r}+1}{\dfrac{1}{r^3}+2}\left(-r^2\dfrac{dy}{dr}\right)+\dfrac{1}{r^4\left(\dfrac{1}{r}+2\right)}y=0$$

$$r^4\dfrac{d^2y}{dr^2}+2r^3\dfrac{dy}{dr}-\dfrac{r^3(r^2+r+1)}{2r^3+1}\dfrac{dy}{dr}+\dfrac{1}{r^3(2r+1)}y=0$$

$$\dfrac{d^2y}{dr^2}+\left(\dfrac{2}{r}-\dfrac{r^2+r+1}{r(2r^3+1)}\right)\dfrac{dy}{dr}+\dfrac{1}{r^7(2r+1)}y=0$$