Question

how do you find the particular solution to 4y''+8y'+2y=(cos^2)(x)

Answer 1

the lead I'm working on is\[\cos^2x=\frac12+\frac12\cos(2x)\]so\[Y_p=A+B\cos(2x)+C\sin(2x)\]but I'm still working on it...

Answer 2

ok

Answer 3

\[Y_p'=-2B\sin(2x)+2C\cos(2x)\]\[Y_p''=-4B\cos(2x)-4C\sin(2x)\]plug it in to the equation:\[4(-4B\cos(2x)-4C\sin(2x))+8(-2B\sin(2x)+2C\cos(2x))+\dots\]\[\dots+2(A+B\cos(2x)+C\sin(2x))\]\[=2A+(-16B+2B+16C)\cos(2x)+(-16B-16C+2C)\sin(2x)\]\[2A=\frac12\to A=\frac14\]\[-14B+16C=\frac12\]\[-16B-14C=0\]I used cramer's rule for this system and got\[B=\frac{-7}{452}\]\[C=\frac2{113}\]so the particular would be\[Y_p=\frac14-\frac7{452}\cos(2x)+\frac2{113}\sin(2x)\]I hope...

Answer 4

wolfram does NOT like this problem by the way
http://www.wolframalpha.com/input/?i=4y%27%27%2B8y%27%2B2y%3Dcos%5E2x
^pretty sure that's not the solution you're looking for

Answer 5

woohoo it's right!!

Answer 6

and yea I know wolfram does not like these types of problems! lol

Answer 7

Awesome!
never done one like that before, this is how I learn :D

Answer 8

that's great!!!