SPECIAL CASE #2 A firstorder differential equation of the form y'=f (axbyc) where b0, can always be reduced to a separable firstorder equation by means of the substitution v=axbyc Example y'=1/ (xy1) Solution If we let v=xy1, then dv/dx=1dy/dx, so the differential equation is transformed into (dv/dx)1=1/v or dv/dx= (1v)/v, so2 solution(s) found x=1 y=0 See steps Step by Step Solution Rearrange Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation (xy)*2(xy)*2(4*x*y)=0 Step by step solution Step 1Let R = ln(u2 v2 w2), u = x 2y, v = 2x − y, and w = 2xy Use the Chain Rule to find ∂R ∂x and ∂R ∂y when x = y = 1 Solution The Chain Rule gives ∂R ∂x = ∂R ∂u ∂u ∂x ∂R ∂v ∂v ∂x ∂R ∂w ∂w ∂x = 2u u2 v 2w ×1 2v u v 2w ×2 2w u v w2 ×(2y) When x = y = 1, we have u = 3, v = 1, and w = 2, so ∂R ∂x = 6 14 ×1 2 14 ×2 4 14 ×2 = 18
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