82_pdfsam_math 54 differential equation solutions odd

82_pdfsam_math 54 differential equation solutions odd -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online