Math 5 (9.10)

September test

Solve y(4)-3y(3)+5y(2)-4y'+4y=0

y=(e^2x)(ax^3+bx^2+cx+d)+fcos(x)+gsin(x)

Solve y(2)-5y(1)+6 y = c1e^-3+c2e^-2

Why, in the proof for solving diff. equations with repeated roots, can we assume that the product of f(t) with e^(r1t) is a solution? -Deanna (I don't know this answer yet.............)

3) Solve: y= -((yx^3+x)/3)y'-yx^2

Solution: (yx^3)/3 + yx + (x^3y^2)/6 = C

-Kelsey

Why are you able to add the two different solutions to form a general solution when you solve a linear second order differential equation? For example, with the equation y' ' - 4y' + 3y = 0, if you get r = 1 and r = 3, why is your general solution y = Ae^(x) + Be^(3x) rather than either y = Ce^(x) or y = Ce^(x)?

You are able to add the two different solutions to the second order linear differential equation because differentiation is linear. If the both the first and solutions satisfy the differential equation and both equal 0, then the addition of the two solutions when plugged into the differential equation will also equal 0. --Amai 12:47, 17 September 2010 (CDT)