# Solving second order linear differential equations with non-constant coefficients [closed]

Can any one tell me what is the general method to solve the second order differential equation like this: $$t(t + 1) y '' + (2 - t^2) y ' - (2 + t) y = (t +1 )^2$$

## closed as off-topic by user26857, user91500, Daniel W. Farlow, Charles, Jack's wasted lifeMar 25 '16 at 13:38

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• I doubt there is a general method here since this isn't an ODE. Whatever method works (if such a method exists) is specific to this particular Diff. EQ. (other than numerical methods which are--or can be--general). – Jared Apr 28 '15 at 7:03
• @Jared : how this isn't this a linear ODE (although with non constant coefficients)? – Tryss Apr 28 '15 at 7:06
• @Tryss I mispoke, I meant ordinary linear differential equations, not ODE's in general. But my original critique stands (imo), either this particular Diff. EQ. has a nice solution or it doesn't--there is no general way to solve such Diff. EQ.s (again, other than numerical methods--of which you can possibly prove whether or not the numerical method can converge). – Jared Apr 28 '15 at 7:17
• @Tryss Well, I guess I'm out of my league here, except my main comment stands: there is no general way to solve such a problem (other than numerical methods). Numerical methods, of which I would consider Taylor or Laurent series as part of. – Jared Apr 28 '15 at 7:27

One rather general method would be to try out a solution in the form of a power series, $$y(t)=\sum_{n=0}^{+\infty} c_n t^n.$$ Insert it and compare the coefficient in front of different powers of $t$.

In this specific case, I'd suggest that you try with the following two functions: 1) $y(t)=t^a$, 2) $y(t)=e^{bt}$. You'll see that they solve the corresponding homogeneous differential equation if $a$ and $b$ are chosen correctly.

Then, to find a particular solution, you might want to try with a polynomial of a certain degree. I'd try a second degree polynomial first. There are methods to decide the particular solution more generally, but in this case I think an ansatz is the easiest way out.

The general solution is of course the sum of the general solution you find out to the homogeneous equation and the particular solution you get.

PS In general, linear differential equations with non-constant coefficients have non-elementary solutions. This is a particular exception.

• To the down voter: Please explain why. Otherwise, how would I be able to improve? – mickep Apr 30 '15 at 16:26

(i) If you know the nontrivial solutions of the homogeneous equation then using "variation of parameters formula" the solutions of inhomogeneous equation can be obtained by integrals.
(ii) If you know one nontrivial solution of the homogeneous equation then using "reduction of order formula" you obtain the other solution by an integral.
In your case the general solution is $$y(t)=\frac{c_1}{t}+c_2 e^t-\frac{t}{2}-1.$$