Question

Which of the following functions is a solution to the differential equation y" + 2y' + y = 0?

Answer 1

Auxiliary: m^2+2m+1=0
m=-1
y = (A+Bx)e^-x

Answer 2

http://www.analyzemath.com/calculus/Differential_Equations/solve_second_order_2.html

Answer 3

Basically your equation is known as a homogeneous solution. So in maths term it is Y subscript H, your y" is actually y^2 and y' is y, so the equation actually is y^2+2y+1=0, solve it you will get -1. with that you know that it is a single and real root. Do note that there are 2 different kinds of roots, (1) single real, (2) real and distinct, (3) complex. single real gives you only 1 value, real and distinct gives you 2, and complex gives you funky valule like roots with i. LOL! thats how i remember it. So basically there is a table that governs each root, single real your general solution should be y = (A+Bx)e^(alpha)x alpha here is basically wad you solved for y just now which is -1. so your solution should look like this y=Ae^-x + Bxe^-x. Go google the other two solution. Shan't spoon feed you too much hahas! Usually if your right hand solution will have a r(x) term you need to "GUESS"... but in this case your r(x) value is 0 so you do not have to do anything. But if there is a r(x) term then you will need to solve it and combine both homogeneous and non homo tgt to get your y(x). Cheers!

Answer 4

would it be y=e^(-t)?

Answer 5

Yep that'd be one of the solutions