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Test 1 Solutions

# Test 1 Solutions - x 2 y 2(2 x 3 y Â 2 y y = 0[6 Solution P...

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MATH 1005B Test 1 Solutions January 28, 2008 [Marks] 1. Solve the initial-value problem 2 y 0 = sin( x ) y ;y (0) = 1. [5] Solution: The equation is separable. 2 yy 0 =s in( x ) ) y 2 = ¡ cos( x )+ c ) y = § p c ¡ cos( x ) : y (0) = 1 ) c =2and y = p 2 ¡ cos( x ). 2. Find the general solution of y 0 + y = 1 y 2 . [5] Solution: The equation is Bernoulli with ® = ¡ 2 :u = y 1 ¡ ® = y 3 ) y = u 1 = 3 ) y 0 = 1 3 u ¡ 2 = 3 u 0 , and the equation becomes 1 3 u ¡ 2 = 3 u 0 + u 1 = 3 = u ¡ 2 = 3 ) u 0 +3 u = 3, which is linear with the integrating factor e 3 x , so the equation becomes ( e 3 x u ) 0 =3 e 3 x ) e 3 x u = e 3 x + c ) u =1+ ce ¡ 3 x ) y =(1+ ce ¡ 3 x ) 1 = 3 . 3. Solve the initial-value problem x 2 y 0 = xy + y 2 ;y (1) = 1. [6] Solution: The equation is y 0 = xy + y 2 x 2 = y x + ³ y x ´ 2 , which is homogeneous. u = y x ) u + xu 0 = u + u 2 ) xu 0 = u 2 ) u ¡ 2 u 0 = 1 x u ¡ 1 =ln j x j + c ) u = ¡ 1 ln j x j + c :y (1) = 1 ) c = ¡ 1 ) u = 1 1 ¡ ln j x j ) y = x 1 ¡ ln j x j . 4. Find the general solution of 2 y 0 +4 y = e ¡ x . [4] Solution: The equation is linear, with the standard form y 0 +2 y = e ¡ x 2 , with the integrating factor e 2 x .Thu s , ¡ e 2 x y ¢ 0 = e x 2 ) e 2 x y = e x 2 + c ) y = e ¡ x 2 + ce ¡ 2
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Unformatted text preview: x 2 y 2 + (2 x 3 y Â¡ 2 y ) y = 0. [6] Solution: P y = 6 x 2 y = Q x , so the equation is exact. f x = P = 3 x 2 y 2 ) f ( x; y ) = x 3 y 2 + g ( y ), and f y = Q ) 2 x 3 y + g ( y ) = 2 x 3 y Â¡ 2 y ) g ( y ) = Â¡ 2 y ) g ( y ) = Â¡ y 2 + c ) f ( x; y ) = x 3 y 2 Â¡ y 2 + c , and the general solution of the equation is x 3 y 2 Â¡ y 2 = k . 6. Find an integrating factor which will make the equation x + y 2 + xyy = 0 exact. Do [4] not solve the equation. Solution: P y = 2 y; Q x = y; P y Â¡ Q x Q = y xy = 1 x ) I I = 1 x ) I ( x ) = x ....
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