Section 4.1

#18. We need to test if cos(2x), 1, & cos^{2}x are linearly independent or dependent. We could set up the Wronskian, though we might recall the half-angle formula for cosine: cos^{2}x = (1/2)(cos(2x)+1), which rearranges to (1)cos^{2}x - (1/2)cos(2x) - (1/2)(1) = 0, so with coefficients 1, -1/2, & -1/2 we get these functions adding up to 0, so we have linearly dependent.

#26. We need to show that e^{x/2} and xe^{x/2} form a fundamental set of solutions for 4y''-4y'+y=0. That means we must show that:

a) e^{x/2} solves the DE. We take 2 derivatives, plug them in to the DE, and we get 0. So it is a solution.

b) xe^{x/2} solves the DE. We take 2 derivatives, plug them in to the DE, and we get 0. So it is a solution.

c) e^{x/2} and xe^{x/2} are linearly independent. We can check the Wronskian of those two functions, and we will find that the Wronskian = e^{x}, which is never zero. So the functions are linearly independent.

Thus, the two functions form a fundamental set of solutions for the DE.

#31. To show that this is the general solution to the NH, we must show that the first two terms form the general solution for the corresponding LH, and the last term is a particular solution to the NH.

LH part: We know (from class) that e^{2x} and e^{5x} are linearly independent. (We could easily show it with a Wronskian, too.) We can also show that both of those functions solve the LH by differentiating them and plugging them into the DE. Since they solve and are lin indep, they form a fundamental set of solutions to the LH, so the complementary solution to the NH is y_{c} = c_{1}e^{2x} + c_{2}e^{5x}.

NH part: We show that 6e^{x} is a solution to the NH by differentiating it and plugging in. It works, so y_{p} = 6e^{x}.

Thus y = c_{1}e^{2x} + c_{2}e^{5x} + 6e^{x} is the general solution to the NH.