1 Exact Equations and Integrating Factors Def Given Mdx+Ndy = 0 with @M@y 6= @N@x , i.e. not exact. If (x;y)M(x;y)dx+ (x;y)N(x;y)dy = 0 is exact, i.e. @ @y( M) = @ @x( N) then (x;y) is called an integrating factor. In example with equation (A), 1t is an integrating factor, in (B) 1ty is an integrating factor. How do we nd integrating factors? What are the conditions on (x;y) so as to make (x;y) an integrating factor? ANS @( M) @y = @( N) @x ; @M @y +M @ @y = @N @x +N @ @x or M @ @y N @ @x + @M @y @N @x = 0 a PDE Is there an integrating factor (x), function of x only? Nd dx + @M @y @N @x = 0 or d = 1 N @M @y @N @x dx ANS YES if ::: is a function of x only. 1 Is there an integrating factor (y), function of y only? Md dy + @M @y @N @x = 0 or d = 1 M @N @x @M @y dy ANS YES if ::: is a function of y only. ) If @M@y 6= @N@x see if (a) 1N @M @y @N @x is a function of x only. YES ) = (x) only. Solve a separable DE for (x). see if (b) 1M @N @x @M @y is a function of y only. YES ) = (y) only. Solve a separable DE for (y). Example dy dx = x2 +y2 +x xy ; y(2) = 1: Find y(1): (x2 +y2 +x)dx+xydy = 0 @M @y = 2y; @N @x = y not exact 1 M @N @x @M @y = 1x2 +y2 +x(y 2y) not a function of y 1 N @M @y @N @x = 1xy(2y y) = 1x Good! 2 d = 1 N @M @y @N @x dx d = 1 xdx)lnj j= lnjxj+ ln ~C ) = Cx is an integrating factor Now DE with = x )(x3 +xy2 +x2)dx+x2ydy = 0 is supposed to be exact CHECK: @M @y = 2xy = @N @x EXACT f(x;y) = x 4 4 + x2y2 2 + x3 3 + (y) f(x;y) = x 2y2 2 + (x) )f(x;y) = x 4 4 + x3 3 + x2y2 2 )d x4 4 + x3 3 + x2y2 2 = 0 ) x 4 4 + x3 3 + x2y2 2 = C 3x4 + 4x3 + 6x2y2 = C1 I.C.)3(2)4 + 4(2)3 + 6(2)2( 1)2 = 104 = C1 3x4 + 4x3 + 6x2y2 = 104 to nd y(1) solve for y y = r 104 x4 4x3 6x2 note the choice of in p y(1) = r104 3 4 6 = 4:02 3