lecture_7 (dragged)

# lecture_7 (dragged) - MA 36600 LECTURE NOTES: WEDNESDAY,...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 28 Linear vs. Nonlinear Solutions Example 1. Consider the initial value problem t y ￿ + 2 y = 4 t2 , y (1) = 2. What is the largest region R such that there exists a unique solution? We solve this problem in three steps: ﬁrst we ﬁnd the general solution to the diﬀerential equation, then we ﬁnd the particular solution using the initial conditions, and ﬁnally we compute the region R by considering properties of this solution. We begin by ﬁnding the general solution to the diﬀerential equation. (Recall that we solved this equation in Lecture 5.) Upon dividing this equation by t, we see that dy + p(t) y = g (t) dt The integrating factor is µ(t) = exp ￿￿ where t 2 , t p(t) = ￿ ￿￿ p(τ ) dτ = exp t g (t) = 4 t. ￿ 2 d τ = t2 . τ Now multiply both sides of the diﬀerential equation by this function: µ(t) dy + µ(t) p(t) y = µ(t) g (t) dt dy t2 + 2 t y = 4 t3 dt d￿2 ￿ t y = 4 t3 =⇒ dt t2 · y (t) = t4 + C for some constant C . Now we can ﬁnd the particular solution to the initial value problem. If we set t = 1 we ﬁnd the equation 2 = 1 + C . This gives C = 1, so that the solution to the initial value problem must be y (t) = t2 + 1 . t2 Note that this solution is deﬁned when t ￿= 0. Finally we compute the region R for which there does exist a unique solution to the initial value problem. We can express the diﬀerential equation y ￿ + p(t) y = g (t) in the form y ￿ = G(t, y ), where G(t, y ) = g (t) − p(t) y = 4 t − 2 y t =⇒ ∂G 2 (t, y ) = − . ∂y t These functions are both continuous when t ￿= 0. The real plane R2 is divided in half by the line t = 0, so our region R must be contained in one of these two. Also, the point (t0 , y0 ) = (1, 2) must be contained in R, so the only possibility is the right-half plane: ￿ ￿ ￿ ￿ 2￿ R = (t, y ) ∈ R ￿ 0 < t < ∞, −∞ < y < ∞ . Example 2. Now consider the initial value problem dy 3 t2 + 4 t + 2 = , dt 2 (y − 1) y (0) = −1. Again, we compute the largest region R such that there exists a unique solution. 1 ...
View Full Document

## This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University-West Lafayette.

Ask a homework question - tutors are online