T1Solutions

T1Solutions - MATH 1005A Test 1 Solutions February 2, 2010...

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MATH 1005A Test 1 Solutions February 2, 2010 [Marks] 1. Solve the initial-value problem 2 y 0 = cos( x ) y ;y (0) = ¡ 2. [5] Solution: 2 yy 0 =cos( x ) ) y 2 =sin( x )+ c ) y = § p sin( x )+ c: y (0) = ¡ 2 2= ¡ p c ) c =4 ) y = ¡ p sin( x )+4. 2. Solve the initial-value problem y 0 = x 2 + y 2 xy ;y ( ¡ 1) = 2. [6] Solution: y 0 = x 2 + y 2 xy = x y + y x ;u = y x ) u + xu 0 = 1 u + u ) xu 0 = 1 u ) uu 0 = 1 x ) 1 2 u 2 =ln j x j + c ) u = § p 2ln j x j + k ) y = § x p 2ln j x j + k: y ( ¡ 1) = 2 ) 2= p k ) k =4 ) y = ¡ x p 2ln j x j +4. 3. Find the general solution of xy 0 +3 y = x +1. [6] Solution: y 0 + 3 x y = x +1 x ) I ( x )= e R 3 x dx = e 3ln j x j = j x j 3 = § x 3 ,andwemaytake I ( x )= x 3 . Then x 3 y 0 +3 x 2 y = x 3 + x 2 ) ( x 3 y ) 0 = x 3 + x 2 ) x 3 y = x 4 4 + x 3 3 + c ) y = x 4 + 1 3 + c x 3 . 4. Find the general solution of x 2 y 3 +( x 3 y 2 +2 y ) y 0 =0 . [6] Solution: P = x 2 y 3 ;Q = x 3 y 2 +2 y; P y =3 x 2 y 2 = Q x ) the equation is exact.
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T1Solutions - MATH 1005A Test 1 Solutions February 2, 2010...

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