The following question was given to as in a recent assignment, namely
Let $$y''+p(t)y'+q(t)y=0~.$$ It is given that $(1+t)^2$ is a solution to this differential equation, and that the wronskian of any two solutions is constant. Find the coefficients of $p$ and $q$.
Say for example the wronskian is a constant, say $C$. Then if $u$ is another solution, then $$u'+ \frac{2u}{(1+t)^2} = C~,$$ from which I get $$u = \frac{Ct+A}{(1+t)^2}~,$$ where $A$ is another constant.
How do I obtain the coefficients $p$ and $q$ ? I could solve simultaneous equations, but then they are messy, as the second derivative of $q$ would involve many terms.
How can Abel's identity for differential equations help me? Another problem is that I don't know if the wronskian is $0$ or not.
But $p$ and $q$ are analytic, so if their wronskian is zero then the two functions are linearly dependent; just putting some facts together.
Thanks, Ben
As DJC pointed out, the differential equation you get by setting the Wronskian to a constant is $$u'(t) = {2 \over 1 + t} u(t) + {C \over (1 + t)^2}$$ This is a first-order linear equation which is solved by the usual methods to $$u(t) = -{C' \over 1 + t} + D(1 + t)^2$$ We can choose any of these to be the second solution, other than a multiple of the first solution $(1 + t)^2$. So we take $u(t) = {1 \over 1 + t}$. To find out what $p(t)$ and $q(t)$ are, you can just plug your two solutions into the original equation and solve for $p(t)$ and $q(t)$. The first one gives $$2(1 + t)p(t) + (1 + t)^2 q(t) = -2$$ The second one gives $$-{1 \over (1 + t)^2}p(t) + {1 \over 1 + t}q(t) = -2{1 \over (1 + t)^3}$$ or equivalently $$-(1 + t)p(t) + (1 + t)^2q(t) = -2$$ Subtract these equations to get $p(t) = 0$. Then solve for $q(t) = -2{1 \over (1 + t)^2}$.
Edit: The Wronskian is actually of the form $Ce^{-\int p(t)}$ by something known as Abel's identity. So if you are willing to use that, you can shorten your work by getting $p(t) = 0$ immediately this way. Then you can just plug in your first solution and solve for $q(t)$.