Reduction of Order problem...

[hank]

The indicated function y1(x) is a solution of the homogeneous equation. Use reduction of order to find a 2nd solution y2(x) of the homogeneous equation and a particular solution of the nonhomogeneous equation.

y" - 4y = 2; y1 = e^-2x

Ok, so....
P(x) = 0
y2 = e^-2x * S dx / e^-4x //Plugging into the formula, the numerator reduces to 1 and the bottom is e^-4x
y2 = e^-2x * (e^4x)/4 //After integration

y2 = (e^2x) / 4
yp = -1/2, by inspection.

However, this is wrong. The book tells me the answer is y2 = e^2x. My question is where does the 1/4 go?

 

[Unco]

G'day Hank,

Although the computation may vary slightly -- such as applying a formula directly as you have, or substituting y2(x) = v(x)*y1(x) for some unknown function v(x) into the DE and so forth -- the aim of the reduction of order method is to obtain a second solution, y2, that is linearly independent to y1. It should be clear that if Cy2 is linearly independent to y1, for some non-zero constant C, then so is y2, and vice versa. Hence your solution is perfectly correct, but you should recognise that so is the book's. You should also note that in your method you have tacitly chosen values for the constants of integrations (plural), which is appropriate for the aim we have described.