This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter 4
We use the superposition principle to ﬁnd a particular solution to the original nonhomogeneous equation. A particular solution, yp,1 (t) to y − 3y + 7y = 7t2 has the form yp,1(t) = A2 t2 + A1 t + A0 . Substitution yields yp,1 − 3yp,1 + 7yp,1 = 2A2 − 3(2A2 t + A1 ) + 7(A2 t2 + A1 t + A0 ) = 7t2 ⇒ ⇒ and so 4 6 t+ . 7 49 The other term in the right-hand side of the original equation is et . A particular solution to yp,1 (t) = t2 + y − 3y + 7y = et has the form yp,2(t) = Aet . Substitution yields yp,2 − 3yp,2 + 7yp,2 = 5Aet = et ⇒ A= 1 5 ⇒ yp,2 (t) = 1t e. 5 (7A2 )t2 + (7A1 − 6A2 )t + (7A0 − 3A1 + 2A2 ) = 7t2 7A2 = 7, 7A1 − 6A2 = 0, 7A0 − 3A1 + 2A2 = 0 ⇒ A2 = 1, A1 = 6/7, A0 = 4/49, By the superposition principle, a general solution to the original equation is y (t) = yh (t) − yp,2(t) + yp,1(t) √ 19t + c2 sin = c1 cos 2 √ 19t 2 e3t/2 − 4 1t 6 e + t2 + t + . 5 7 49 23. The corresponding homogeneous equation in this problem is similar to that in Problem 13. Thus, y1 (t) = cos 4θ and y2 (t) = sin 4θ are its two linearly independent solutions, and a general solution is given by yh (θ) = c1 cos 4θ + c2 sin 4θ . For a particular solution to the original equation, we use the variation of parameters method. Letting yp (θ) = v1 (θ) cos 4θ + v2 (θ) sin 4θ, 250 ...
View Full Document