Unformatted text preview: Chapter 2
and multiplying (2.18) by µ(xy ), we get a diﬀerential equation µ(xy )M (x, y )dx + µ(xy )N (x, y )dy = 0. Let us check it for exactness. First we note that µ (z ) = exp H (z )dz = exp H (z )dz H (z )dz = µ (z )H (z ). (2.20) Next, using this fact, we compute partial derivatives of the coeﬃcients in (2.20). ∂M (x, y ) ∂ ∂ (xy ) {µ(xy )M (x, y )} = µ (xy ) M (x, y ) + µ(xy ) ∂y ∂y ∂y ∂M (x, y ) = µ(xy )H (xy ) xM (x, y ) + µ(xy ) ∂y ∂M (x, y ) = µ(xy ) H (xy ) xM (x, y ) + , ∂y ∂N (x, y ) ∂ (xy ) ∂ {µ(xy )N (x, y )} = µ (xy ) N (x, y ) + µ(xy ) ∂x ∂x ∂x ∂N (x, y ) = µ(xy )H (xy ) yN (x, y ) + µ(xy ) ∂x ∂N (x, y ) = µ(xy ) H (xy ) yN (x, y ) + . ∂x But (2.19) implies that ∂M ∂N − = (xM − yN )H (xy ) ∂x ∂y and, therefore, ∂ [µ(xy )M (x, y )] ∂ [µ(xy )N (x, y )] = . ∂y ∂x This means that the equation (2.20) is exact. 17. (a) Expressing the family y = x − 1 + ke−x in the form (y − x + 1)ex = k , we have (with notation of Problem 32) F (x, y ) = (y − x + 1)ex . We compute ∂F ∂ ∂ (y − x + 1) x d(ex ) = [(y − x + 1)ex ] = e + (y − x + 1) ∂x ∂x ∂x dx = −ex + (y − x + 1)ex = (y − x)ex , 78 ⇔ yNH (xy ) + ∂N ∂M = xMH (xy ) + , ∂x ∂y ...
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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 Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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