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Solution Techniques for Elementary Partial Differential Equations 3rd Edition
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Solution Techniques for Elementary Partial Differential Equations 3rd Edition, Instructor’s Solutions Manual Solution Techniques for Elementary Partial Differential Equations 3rd Edition, Instructor’s Solutions Manual Solution Techniques for Elementary Partial Differential Equations, Third Edition, Instructor’s Solutions Manual ch1-ch14 Chapter 1 Section 1.1 This is a variables separable equation and y = 0 is a solution. If y 6= 0, then dy y = 2x x 2 + 1 dx ⇒ ln|y| = ln(x 2 + 1) + C 0 ⇒ y(x) = C(x 2 + 1), where C 6= 0. The value C = 0 generates the singular solution y = 0. Proceeding as in 1, for y 6= −1 we have y ′ = 3x 2 (y + 1) ⇒ dy y + 1 = 3x 2 ⇒ ln|y + 1| = x 3 + C 0 ⇒ y = Ce x 3 − 1, where C 6= 0. The value C = 0 generates the singular solution y = −1. This is a linear equation, solved by the integrating factor method: dy dx + 2 x − 1 y = x x − 1 ⇒ µ = exp ?Z 2 x − 1 dx ? = e 2ln|x−1| = (x − 1) 2 ⇒ y(x) = 1 (x − 1) 2 Z (x − 1) 2 x x − 1 dx = 1 (x − 1) 2 Z (x 2 − x)dx = (x − 1) −2 ( 1 3 x 3 − 1 2 x 2 + C ) . Proceeding as in 3, we have dy dx − 2 x y = x 3 e x ⇒ µ = exp ?Z − 2 x dx ? = e −lnx 2 = 1 x 2 ⇒ y(x) = x 2 Z 1 x 2 x 3 e x dx = x 2 Z xe x dx = x 2 ? xe x − Z e x dx ? = x 2 (xe x − e x + C). download via https://r.24zhen.com/n33WM
