Homework1Solns

# Homework1Solns - SHORT CALCULUS Math 16C Sec 1 Fall 2008 Homework#1 Solutions Peter Malkin Section C.1 Question 28 Since y = x(ln x C we have dy dx

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Unformatted text preview: SHORT CALCULUS Math 16C Sec 1 Fall 2008 Homework #1 Solutions Peter Malkin Section C.1 Question 28 Since y = x (ln x + C ), we have dy dx = (ln x + C ) + x ( 1 x ) = ln x + C + 1. Thus, x + y- xy ′ = x + [ x (ln x + C )]- x [ln x + C + 1] = x + x ln x + Cx- x ln x- Cx- x = 0 as required. Question 46 Since y = Ce x − x 2 , we have y ′ = C (1- 2 x ) e x − x 2 . y ′ + (2 x- 1) y = [ C (1- 2 x ) e x − x 2 ] + (2 x- 1)[ Ce x − x 2 ] = C (1- 2 x ) e x − x 2- C (1- 2 x ) e x − x 2 = 0 as required. Then, since y = 2 when x = 1, we have 2 = Ce 1 − 1 2 = C ⇔ C = 2. So, the particular solution is y = 2 e x − x 2 . 1 Section C.2 Question 10 dy dx = x 2 y ⇔ 1 y dy = x 2 dx ⇔ integraldisplay 1 y dy = integraldisplay x 2 dx ⇔ ln | y | = 1 3 x 3 + C ( C is a constant) ⇔ | y | = Ae x 3 3 ( A is a constant) ⇔ y = Ae x 3 3 The general solution is y = Ae x 3 3 . Question 22 dy dx- y ( x + 1) = 0 ⇔ dy dx = y ( x + 1) ⇔ 1 y dy = ( x + 1) dx ⇔ integraldisplay 1 y dy = integraldisplay ( x + 1) dx ⇔ ln | y | = 1 2 x 2 + x + C ( C is a constant) ⇔ | y | = e ( 1 2 x 2 + x + C ) ⇔ y = Ae ( 1 2 x 2 + x ) ( A is a constant) The general solution is y = Ae ( 1 2 x 2 + x ) ....
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## This note was uploaded on 01/25/2011 for the course MAT 16C taught by Professor Kouba during the Spring '08 term at UC Davis.

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Homework1Solns - SHORT CALCULUS Math 16C Sec 1 Fall 2008 Homework#1 Solutions Peter Malkin Section C.1 Question 28 Since y = x(ln x C we have dy dx

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