exam1_ODE_solution

# exam1_ODE_solution - Solutions of Exam 1, MAP 2302,...

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Unformatted text preview: Solutions of Exam 1, MAP 2302, Fall’11 1 . One has y prime = f ( x, y ) = 4 x √ y- 1. So f is continuous for y ≥ 1, but ∂f/∂y = 2 x/ √ y- 1 does not exist at y = 1. Since f ( x, 1) = 0 for all x , the equation has a solution y ( x ) = 1. If y > 1, then by separating the variables one infers another solution: integraldisplay dy √ y- 1 = integraldisplay 4 xdx + C 1 = ⇒ 2 radicalBig y- 1 = 2 x 2 + C 1 = ⇒ y = 1 + ( x 2 + C ) 2 where C = C 1 / 2. 2 . Put Ndx + Mdy = 0 where N = 3 x 2 y + 2 x and M = x 3 + 3 y 3 . The equation is exact if ∂N/∂y = ∂M/∂x . One has ∂N/∂y = 3 x 2 and ∂M/∂x = 3 x 2 , i.e. the equation is exact. Its general solution is F ( x, y ) = C where ∂F/∂x = N and ∂F/∂y = M . It follows from the first equation that F = integraltext (3 x 2 y + 2 x ) dx = x 3 y + x 2 + h ( y ). Therefore ∂F/∂y = x 3 + h prime ( y ) which is then substituted into the second equation to obtain h prime ( y ) = 3 y 2 and h ( y ) = y 3 . So F ( x, y ) = x 3 y + x 2 + y 3 and the general solutions reads: x 3 y +...
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## This note was uploaded on 12/15/2011 for the course MAP 2302 taught by Professor Tuncer during the Spring '08 term at University of Florida.

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exam1_ODE_solution - Solutions of Exam 1, MAP 2302,...

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