Differntial Equation (Part 2) Free download as PDF File (pdf), Text File (txt) or read online for free If integrating factor of x(1 x 2)dy (2x 2 y y ax 3)dx = 0 is e ∫Pdx, then P is equal to (A) (2x 2 ax 3)/x(1 x 2) (B) (2x 2 1) (2x 2 1)/ax 3 (D) (2x 2 1)/x(1 x 2) = 3(x dx)x 2 2x(dx) (dx)2 – 3x 3 menentukan dy bagi = 3x 3 2x 2(dx) x(dx)2 x 2(dx) 2x(dx)2 (dx)3 – 3x 3 = 3x 3 3x 2(dx) 3x(dx)2 (dx)3 – 3x 3 dx
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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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The graph of y=(x3)^2(x4)^3 represents a graph of a polynomial function On the given graph you can find all of the important points for function y=(x3)^2(x4)^3 (if they exist)Which statement about the relationship between these two graphs is true? Graph #y2=2/3(x4)# The easiest way to find points on the line is to convert the given equation in point slope form to slope intercept form #y=mxb#, where #m# is the slope, and #b# is the yintercept In order to do this, solve the point slope equation for #y# #y2=2/3(x4)# Add #2# to both sides #y=2/3(x4)2# Simplify #2/3(x4)# to #(2(x4))/3#
Answered Identify The Function Shown In This Bartleby
