350_pdfsam_math 54 differential equation solutions odd

350_pdfsam_math 54 differential equation solutions odd -...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 6 To satisfy the initial conditions, first we find 1 + C1 + C2 (ln x + 1) + C3 (ln x)2 + 2 ln x , x 2 ln x 2 1 C2 y (x) = − 2 + + C3 + . x x x x y (x) = Substituting the initial conditions, y (1) = 3, y (1) = 3, and y (1) = 0, we get the system 3 = y (1) = C1 , 3 = y (1) = 1 + C1 + C2 , 0 = y (1) = −1 + C2 + 2C3 Thus, y (x) = ln x + 3x − x ln x + x(ln x)2 is the desired solution. 23. Substituting y1 (x) = sin x and y2 (x) = x into the given differential operator yields L[sin x] = (sin x) + (sinx) + x(sin x) = − cos x + cos x + x sin x = x sin x, L[x] = (x) + (x) + x(x) = 0 + 1 + x2 = x2 + 1. Note that L[y ] is a linear operator of the form (7). So, we can use the superposition principle. (a) Since 2x sin x − x2 − 1 = 2(x sin x) − (x2 + 1), by the superposition principle, y (x) = 2y1(x) − y2 (x) = 2 sin x − x is a solution to L[y ] = 2x sin x − x2 − 1. (b) We can express 4x2 + 4 − 6x sin x = 4(x2 + 1) − 6(x sin x). Hence, y (x) = 4y2 (x) − 6y1 (x) = 4x − 6 sin x is a solution to L[y ] = 4x2 + 4 − 6x sin x. 346 ⇒ C1 = 3, C1 + C2 = 2, C2 + 2C3 = 1 ⇒ C1 = 3, C2 = −1, C3 = 1. ...
View Full Document

Ask a homework question - tutors are online