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Variation of Parameters
Notice there the differential equation,
y′′ + q (t) y′ + r (t) y = g (t)
Suppose that y1(t) and y2(t) are a fundamental set of solutions for
y′′ + q (t ) y′ + r (t ) y = 0
Depending on the problem and the person, some will determine the formula easier to notice and use, whereas others will determine the process used to find the formula easier. The illustrations in this section will be done using the formula.
Before proceeding along with a couple of illustrations let's first address the issues including the constants of integration which will arise out of the integrals. Placing in the constants of integration will provide the following.
The last quantity in the parenthesis is nothing more than the complementary solution along with c1 = - c and c2 = k and we identify that if we plug this in the differential equation this will simplify out to zero as this is the solution to the homogeneous differential equation. Conversely, these terms add nothing to the particular solution and thus we will go ahead and suppose that c = 0 and k = 0 in all the illustrations.
One last note before we proceed along with illustrations. Do not worry about that of your two solutions in the complementary solution is y1(t) and that one is y2(t). This doesn't matter. You will finds out the same answer no matter that one you select to be y1(t) and which one you choose to be y2(t).
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4.4238/[1.047+{1.111*[9.261/7.777]}*1.01
I've termed this section as Intervals of Validity since all of the illustrations will involve them. Though, there is many more to this section. We will notice a couple of theorems
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