1. ## Variation of Parameters

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.

2. Originally Posted by fudawala
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: $y'' + p(t)y' + q(t) = f(t)$ having homogeneous solutions $y_1(t)$ and $y_2(t)$, we assume, by the method of Variation of Parameters, that a particular solution exists of the form:

$y_p = A(t)y_1(t) + B(t)y_2(t)$

where $A(t)$ and $B(t)$ are found from the system of equations:

$A'(t)y_1 + B'(t)y_2 = 0$ ...............(1)
$A'(t)y_1' + B'(t)y_2' = f(t)$ ...............(2)

since $y_1$ and $y_2$ are knowns, we can find the values of $A'(t)$ and $B'(t)$ by Cramer's Rule, and then integrate them to get $A(t)$ and $B(t)$

then as always, our general solution $y_g(t)$ is given by:

$y_g(t) = y_1(t) + y_2(t) + y_p(t)$

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