{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

57_pdfsam_math 54 differential equation solutions odd

# 57_pdfsam_math 54 differential equation solutions odd - x...

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

Exercises 2.3 (d) We want the value of the initial point for the solution in part (c) to be the value of the solution found in part (b) at the point x = 2. This value is given by y (2) = 2 1+2 e 2 =1+2 e 2 . Thus the initial point we seek is y (2) = 1 + 2 e 2 . Using this initial point to Fnd the constant C g iveninpart(c)y ie lds 1+2 e 2 = y (2) = 2 3 1 9 + Ce 6 C = 4 9 e 6 +2 e 4 . Thus, the solution of the equation on the interval x> 2isg ivenby y = x 3 1 9 + ± 4 9 e 6 +2 e 4 ² e 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x ≥ 0: y = x − 1 + 2 e − x , ≤ x ≤ 2; x 3 − 1 9 + ³ 4 9 e 6 + 2 e 4 ´ e − 3 x , 2 < x. (e) The graph of the solution is given in ±igure B.18 of the answers in the text. 33. (a) Writing the equation in standard form yields dy dx + 2 x y = 3 . Therefore, P ( x ) = 2 /x and µ ( x ) = exp ³Z 2 x dx ´ = exp (2 ln | x | ) = | x | 2 = x 2 . Hence d dx ( x 2 y ) = 3 x 2 ⇒ x 2 y = Z 3 x 2 dx = x 3 + C ⇒ y = x + C x 2 53...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online