70_pdfsam_math 54 differential equation solutions odd

70_pdfsam_math 54 differential equation solutions odd - x )...

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Chapter 2 Integrating both sides yields Z x sin xdx = Z y 1 y 2 dy x ( cos x ) Z ( cos x ) dx = Z ± 1 y 1 y 2 ² dy ⇒− x cos x +sin x =ln | y | + 1 y + C, where we applied integration by parts to fnd R x sin xdx . Substitution oF the initial condition, y ( π )=1,results π cos π +sin π =ln | 1 | + 1 1 + C C = π 1 . So, the solution to the initial value problem is x cos x +sin x =ln y +1 /y + π 1 . (Since y ( π )=1 > 0, we have removed the absolute value sign in the logarithmic term.) 27. (a) We want to fnd M ( x, y )sothatFor N ( x, y )=sec 2 y x/y we have M y ( x, y )= N x ( x, y )= 1 y . ThereFore, we must integrate this last expression with respect to y .Tha ti s , M ( x, y )= Z ± 1 y ² dy = ln | y | +
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Unformatted text preview: x ) , where f ( x ), the “constant” oF integration, is a Function only oF x . (b) We want to fnd M ( x, y ) so that For N ( x, y ) = sin x cos y − xy − e − y we have M y ( x, y ) = N x ( x, y ) = cos x cos y − y. ThereFore, we must integrate this last expression with respect to y . That is M ( x, y ) = Z (cos x cos y − y ) dy = cos x Z cos y dy − Z y dy = cos x sin y − y 2 2 + f ( x ) , where f ( x ), a Function only oF x , is the “constant” oF integration. 66...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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