349_pdfsam_math 54 differential equation solutions odd

# 349_pdfsam_math 54 differential equation solutions odd -...

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: Exercises 6.1 (b) To ﬁnd the solution that satisﬁes the initial conditions, we must diﬀerentiate the general solution y (x) twice with respect to x. Thus, we have y (x) = C1 ex − C2 e−x cos 2x − 2C2 e−x sin 2x − C3 e−x sin 2x + 2C3 e−x cos 2x + 2x = C1 ex + (−C2 + 2C3 ) e−x cos 2x + (−2C2 − C3 ) e−x sin 2x + 2x , y (x) = C1 ex + (C2 − 2C3 ) e−x cos 2x − 2 (−C2 + 2C3 ) e−x sin 2x − (−2C2 − C3 ) e−x sin 2x + 2 (−2C2 − C3 ) e−x cos 2x + 2 = C1 ex + (−3C2 − 4C3 ) e−x cos 2x + (4C2 − 3C3 ) e−x sin 2x + 2 . Plugging the initial conditions into these formulas, yields the equations y (0) = C1 + C2 = −1, y (0) = C1 − C2 + 2C3 = 1, y (0) = C1 − 3C2 − 4C3 + 2 = −3. By solving these equations simultaneously, we obtain C1 = −1, C2 = 0, and C3 = 1. Therefore, the solution to the initial value problem is given by y (x) = −ex + e−x sin 2x + x2 . 21. In the standard form, given equation becomes y+ 1 1 3 − ln x y − 3y= . 2 x x x3 Since its coeﬃcients are continuous on (0, ∞), we can apply Theorems 2 and 4 to conclude that a general solution to the corresponding homogeneous equation is yh (x) = C1 x + C2 x ln x + C3 x(ln x)2 and a general solution to the given nonhomogeneous equation is y (x) = yp (x) + yh (x) = ln x + C1 x + C2 x ln x + C3 x(ln x)2 . 345 ...
View Full Document

## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

Ask a homework question - tutors are online