SKEDSOFT
Second Order ODE with Variable Coefficients:
Let us now return to the more general 2nd order, linear ODE
y’’ p(x) y’ q(x) y = g(x)
How do we find a complementary function and particular integral in this case?
Frobenius Series Solution of Ordinary Differential Equations:
At the start of the Differential equation section of the 1B21 course last year, you met the linear first-order seperable equations:
where a is constant . You were also shown how to integrate the equation to get the solution,
Where A is an arbitrary constant. The solution can be expanded in a power series in x and I want to show explicity that this power series does indeed satiesfy the equations.
Notice that although similar terms are pushed one to right, y' is clearly equal to αy. Let us know see how this is handled using the summation notation. Here,
Above equation does not yet look like α times. Above equation because one sees xn-1 rather than xn . This is reflection of all the term being pushed one over, as noted in the long hand representation above. To show the equivalence first note the n = 0 term is actually absent from the series for y' and so the series effectively starts at n = 1. Secondly, the n in the series is a dummy variable, like an integer variable, which does not occur in the final answer. we can therefore make the substitution m = n -1 in above equation to find.
Which is exactly αy. Generally the boot is on the other foot. We often end up with a different equation which we can not solve by the inspection as we have done here. we want then to develop techniques for finding directly series solution for differential equation. Let us try for a solution of above equation in the form,
where a0 ≠ 0. Thus the value of the index k is defined by the condition that the first term a0 does not vanish. Differentiating term-by-term leads to:
Inserting the above two equations into the different equation, we find:
