The question to this problem is t^2y'' - 2y = 3t^2 - 1, t > 0; y1(t) = t^2, y2(t) = t^-1
Where y1 and y2 satisfy the corresponding homogeneous equation. Then the particular solution to that problem would be:
At^2 + Bt + C right.
The question to this problem is t^2y'' - 2y = 3t^2 - 1, t > 0; y1(t) = t^2, y2(t) = t^-1
Where y1 and y2 satisfy the corresponding homogeneous equation. Then the particular solution to that problem would be:
At^2 + Bt + C right.
No, you do not guess solutions with Variation of Parameters as you do with Undetermined coefficients.
Here's the method of Variation of Parameters:
Given a differential equation of the form: having homogeneous solutions and , we assume, by the method of Variation of Parameters, that a particular solution exists of the form:
where and are found from the system of equations:
...............(1)
...............(2)
since and are knowns, we can find the values of and by Cramer's Rule, and then integrate them to get and
then as always, our general solution is given by:
Most textbooks, including your own, condenses this whole method into a single formula involving an integral, look it up if you can't bother with Cramer's rule