Difference between revisions of "Math 5 (9.10)"

Revision as of 13:11, 17 September 2010

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? Answer: We actually can't assume. This method of assuming is called reduction of order and is more like an educated guess of what the answer is that is later verified. --Deanna (crappy answer I know, but it's the best I can come up with)

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)

Can you solve any second order differential equations using separation of variables? Try y^(2)-1 = 0, using both methods. Do you get the same answer? Scott--Swise 12:48, 17 September 2010 (CDT)

Prove whether or not the following differential equation is exact or not, and then solve it:

(5x^2y + y)dx + ((5/x)x^3 + x)dy = 0 Solution:

-Yes, the equation is exact.

-F=(5/3)x^3 + x + C

--Jeallen 12:55, 17 September 2010 (CDT)

Solve: 6y(4)-5y(3)+y(2)=0 y=c1e^(x/2)+c2e^(x/3)+c3+c4x

myles--Mlewis 12:58, 17 September 2010 (CDT)

Solve:3y(2)+18iy-27=0 Still working on solution!