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math1005ass1sol

# math1005ass1sol - Total of 40 marks MATH1005A 1(solutions...

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Total of 40 marks MATH1005A, Summer2009Assignment 1 (solutions) Problem 1 . Show that if ( Q x - P y ) /P is a function of y alone, then the equation I y P + IP y = I x Q + IQ x (1) enables us to find an integrating factor for P + Qy 0 = 0 ( 4 marks ). Use this method to solve the equation y - 6 x 2 y 3 + (2 x - 8 x 3 y 2 ) y 0 = 0 . (8 marks) Solution: If ( Q x - P y ) /P is a function of y alone then any solution I ( y ) to the separable equation dI dy = I ( Q x - P y ) P , is also a solution to (1) since I x = 0 in this case. Recall that (1) holds if and only if the new equation IP + ( IQ ) y 0 = 0 is exact, i.e. I is an integrating factor for (1). Note that P = y - 6 x 2 y 3 and Q = 2 x - 8 x 3 y 2 . Thus Q x = 2 - 24 x 2 y 2 and P y = 1 - 18 x 2 y 2 . Then ( Q x - P y ) /P = 1 /y . So one has to solve the equation: dI/I = dy/y, to find the integrating factor I . It is easily seen that I ( y ) = y is a solution to this equation. So the new equation IP + ( IQ ) y 0 = 0, is y 2 - 6 x 2 y 4 + (2 xy - 8 x 3 y 3 ) y 0 = 0 . Therefore f ( x, y ) = R ( IP ) dx = y 2 x - 2 x 3 y 4 + g ( y ) and 2 xy - 8 x 3 y 3 = IQ = f y = 2 xy - 8 x 3 y 3 + g 0 ( y ) , 1

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math1005ass1sol - Total of 40 marks MATH1005A 1(solutions...

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