Math1B Quiz5Solutions

Math1B Quiz5Solutions - 1 3 = 0 + 0 + C. Then C = 1 3 1 3 y...

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QUIZ 5 SOLUTION Question 1 dy dx = x + ln( x ) + xy + ln( x ) y = ( x + ln( x ))( y + 1) . dy dx 1 y + 1 = x + ln( x ) . Z 1 y + 1 dy = Z ( x + ln( x )) dx ln | y + 1 | = x 2 2 + x ln( x ) - x + C. | y + 1 | = e x 2 2 + x ln( x ) - x + C = Ke x 2 2 + x ln( x ) - x Since y > - 1 then | y + 1 | = y + 1 and so y = Ke x 2 2 + x ln( x ) - x - 1 . Question 2 dy dx = x 2 y 2 + 1 y 2 = x 2 + 1 y 2 . dy dx y 2 = x 2 + 1 . Z y 2 dy = Z ( x 2 + 1) dx. 1 3 y 3 = 1 3 x 3 + x + C. From initial condition
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Unformatted text preview: 1 3 = 0 + 0 + C. Then C = 1 3 1 3 y 3 = 1 3 x 3 + x + 1 3 . Multiply by 3 and then take cubic root y = 3 p x 3 + 3 x + 1 . 1...
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This note was uploaded on 07/10/2008 for the course MATH 1B taught by Professor Reshetiken during the Spring '08 term at Berkeley.

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