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Unformatted text preview: MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 4 Exact Equations (cont’d) Exact Equations. Say that we have a first order differential equation in the form M ( x,y ) dx + N ( x,y ) dy = 0 . Imagine for the moment that we could find a function z = f ( x,y ) such that dz = ∂f ∂x dx + ∂f ∂y dy = M ( x,y ) dx + N ( x,y ) dy. Then dz = 0 i.e., f ( x,y ) = C must be a constant. This would be an implicit solution to the differential equation. The constant C can be found by specifying that a point ( x ,y ) be on the integral curve f ( x,y ) = C . To recap, if we can find a function z = f ( x,y ) such that ∂f ∂x = M ( x,y ) and ∂f ∂y = N ( x,y ) then we say the first order equation is an exact equation . This terminology is short-hand for saying “the differential equation is exactly the differential of a function.” Example. Consider the differential equation dy dx =- 2 x + y 2 2 xy . (This equation is neither linear nor separable.) Multiply both sides by N ( x,y ) = 2 xy to find the differential equation ( 2 x + y 2 ) dx + (2 xy ) dy = 0 . This is an exact equation because if we choose the function f ( x,y ) = x 2 + xy 2 = ⇒ ∂f ∂x = 2 x + y 2 , ∂f ∂y = 2 xy. Hence the solution to the differential equation is f ( x,y ) = C i.e., x 2 + xy 2 = C for some constant C . Criteria for Exact Equations. We will show that the first order differential equation M ( x,y ) dx + N ( x,y ) dy = 0 is an exact equation if and only if ∂M ∂y = ∂N ∂x . (Note how M = M ( x,y ) is the function multiplied by the differential dx , yet we take the partial with respect to y !) This gives a quick criteria to check that the equation is exact.!...
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This note was uploaded on 05/24/2011 for the course MA 36600 taught by Professor Ma during the Spring '09 term at Purdue.
- Spring '09