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Unformatted text preview: r and θ , M ( r, θ ) = θ , and N ( r, θ ) = 3 r − θ − 1. Because ∂M ∂θ = 1 6 = 3 = ∂N ∂r , the equation is not exact. With r as the dependent variable, the equation takes the form dr dθ = − 3 r − θ − 1 θ = − 3 θ r + θ + 1 θ , and it is linear. Since the righthand side in the above equation cannot be factored as p ( θ ) q ( r ), the equation is not separable. 9. We have that M ( x, y ) = 2 xy + 3 and N ( x, y ) = x 2 − 1. Therefore, M y ( x, y ) = 2 x = N x ( x, y ) and so the equation is exact. We will solve this equation by ±rst integrating M ( x, y ) with respect to x , although integration of N ( x, y ) with respect to y is equally easy. Thus F ( x, y ) = Z (2 xy + 3) dx = x 2 y + 3 x + g ( y ) . 60...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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