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 x sin x dx = y 1 y 2 dy x ( cos x ) ( cos x ) dx = 1 y 1 y 2 dy x cos x + sin x = ln | y | + 1 y + C, where we applied integration by parts to find x sin x dx . 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 find M ( x, y ) so that for 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 . That is, M ( x, y ) = 1 y dy = ln | y | + f
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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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