Aerith's Knight
11-13-2008, 11:05 PM
Ok, I have an exam tomorrow, and I was just practising, and I can't get this freaking differential equation. >:[
This is what I have so far:
x''(t) -2x'(t) + 5x(t) = 4t * e<sup>t</sup>
Homogenous:
(a<sup>2</sup> - 2a + 5) e<sup>at</sup> = 0
a<sup>2</sup> - 2a + 5 = 0
D = (-2)<sup>2</sup> - 4 * 5 * 1 = -16
a = (2 +/- 4i) / 2 = 1 +/- 2i
x(t) = Ae<sup>(1+2i)t</sub> + Be<sub>(1-2i)t</sub>
Re(x(t)): e<sup>(1+2i)t</sup> = e<sup>t</sup> * (cos(2t) - i sin(2t))
= e<sup>t</sup> * cos(2t) - i * e<sup>t</sup> sin(2t)
(1+2i)* = 1-2i
x(t) = Ae<sup>(1+2i)t</sup> + Be<sup>(1-2i)t </sup>
= Ae<sup>t</sup> * cos(2t) - Ai * e<sup>t</sup> sin(2t) + Be<sup>t</sup> * cos(2t) + Bi * e<sup>t</sup> sin(2t)
x(t) = (A+B)cos(2t) * e<sup>t</sup> + (B-A)i * sin(2t) * e<sup>t</sup>
Can someone help me out with the particular solution?
This is what I have so far:
x''(t) -2x'(t) + 5x(t) = 4t * e<sup>t</sup>
Homogenous:
(a<sup>2</sup> - 2a + 5) e<sup>at</sup> = 0
a<sup>2</sup> - 2a + 5 = 0
D = (-2)<sup>2</sup> - 4 * 5 * 1 = -16
a = (2 +/- 4i) / 2 = 1 +/- 2i
x(t) = Ae<sup>(1+2i)t</sub> + Be<sub>(1-2i)t</sub>
Re(x(t)): e<sup>(1+2i)t</sup> = e<sup>t</sup> * (cos(2t) - i sin(2t))
= e<sup>t</sup> * cos(2t) - i * e<sup>t</sup> sin(2t)
(1+2i)* = 1-2i
x(t) = Ae<sup>(1+2i)t</sup> + Be<sup>(1-2i)t </sup>
= Ae<sup>t</sup> * cos(2t) - Ai * e<sup>t</sup> sin(2t) + Be<sup>t</sup> * cos(2t) + Bi * e<sup>t</sup> sin(2t)
x(t) = (A+B)cos(2t) * e<sup>t</sup> + (B-A)i * sin(2t) * e<sup>t</sup>
Can someone help me out with the particular solution?