- #1
checkmatechamp
- 23
- 0
Homework Statement
y'' + 4y = t2 + 6et; y(0) = 0; y'(0) = 5
Homework Equations
The Attempt at a Solution
So, getting the general solution, we have r2 + 4 = 0, so r = +/- 2i
So the general solution is yc = sin(2t) + cos(2t)
I then used the method of undetermined coefficients to figure that the particular solution had a form At2 + Bt + C + Det + E = 0, and found that A = 0.25, B = 0 , C = 0, D = 1.2, E = 2
So the overall solution is y = c1sin(2t) + c2cos(2t) + 0.25t2 + 1.2et + 2
The derivative is y' = 2c1*cos(2t) - 2c2*sin(2t) + 0.5t + 1.2et
So I got that 0 = 0 + c2 + 0 + 1.2 + 2, which means that c2 + 3.2 = 0, which means that c2 = -3.2
5 = 2c1 - 0 + 0 + 1.2, which means that 2c1 = 3.8, and c1 = 1.9
So the overall equation should be y = 1.9sin(2t) - 3.2*cos(2t) + 0.25t2 + 1.2et + 2.
But when I enter it as an answer, it tells me it's wrong. Where did I make a mistake?