math1005test1sol

# math1005test1sol - Total of 40 marks MATH1005A, Summer2009...

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Unformatted text preview: Total of 40 marks MATH1005A, Summer2009 Test 1 (Solutions) Problem 1 (8 marks) . Find a solution to the following separable initial value prob- lem: du dt = 2 t + 1 2( u- 1) , u (0) =- 1 . You need to solve for u ( t ) explicitly to get the full mark. Solution: From the equation we get 2( u- 1) du = (2 t + 1) dt . So Z 2( u- 1) du = Z (2 t + 1) dt. Hence u 2- 2 u = t 2 + t + c , for some constant c . So u 2- 2 u- ( t 2 + t + c ) = 0. So u = 2 ± p 4 + 4( t 2 + t + c ) 2 = 1 ± √ t 2 + t + c + 1 . Since u (0) =- 1, one obtains c = 3 and u = 1- √ t 2 + t + 4 . Problem 2 (8 marks) . Show that the first order ODE: dy dx = x 2 + y 2 xy , is homogeneous. Solve the equation. Solution: Note that x 2 + y 2 xy = 1 + ( y/x ) 2 y/x = 1 + v 2 v = g ( v ) . Thus Z vdv = Z dv g ( v )- v = Z dx x . So v 2 / 2 = ln | x | + c . So v = ± p 2 ln | x | + c. Finally y = ± x p 2 ln | x | + c. Total of 40 marks MATH1005A, Summer2009 Test 1 (Solutions) Problem 3 (8 marks) . Solve the following first order ODE:...
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## This note was uploaded on 01/18/2012 for the course MATH 1005 taught by Professor Any during the Summer '07 term at Carleton CA.

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math1005test1sol - Total of 40 marks MATH1005A, Summer2009...

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