MATH 4B Midterm 1 Solutions

# Solution we have m x y 2 xy 3 x 2 and n x y x 2 2 y

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Solution: We have M ( x, y ) = 2 xy - 3 x 2 and N ( x, y ) = x 2 - 2 y , then ∂M ∂y = 2 x = ∂N ∂x . Thus the equation is exact. (b) Find the solution to (**) satisfying the initial condition y (1) = 1. Solution: Since the equation is exact we know that all the solutions are of the form Ψ( x, y ) = c, where Ψ( x, y ) is a function satisfying Ψ ∂x = M, Ψ ∂y = N. Thus Ψ( x, y ) = Z M ( x, y ) dx = Z 2 xy - 3 x 2 dx = x 2 y - x 3 + C ( y ) , where C ( y ) is a function depending only on the variable y . To find C ( y ) we take the partial derivative of Ψ( x, y ) with respect to y and compare with N ( x, y ). We obtain Ψ ∂y = x 2 + C 0 ( y ) N ( x, y ) = x 2 + C 0 ( y ) x 2 - 2 y = x 2 + C 0 ( y ) - 2 y = C 0 ( y ) - y 2 + c = C ( y ) . Therefore the solutions to the exact equation are of the form x 2 y - x 3 - y 2 = c. Since y (1) = 1 we must have (1) 2 (1) - (1) 3 - (1) 2 = c and so the solution to the IVP is given by x 2 y - x 3 - y 2 = - 1 . Page 6

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