Since y = v1/3, y = 3 r C x2 3x (f) y0ycot x = cos x Solution We see that this equation is already linear, with p(x) = cot x = cos x sin x, so its integrating factor is m(x) = e R cot xdx = elnjsin xj = sin x Multiplying by this integrating factor, we get (ysin x)0= sin xcos x, so integrating gives ysin x = 1 2 sin 2 x C, and y = 1 2 sin xX dy/dx y = 4x 1 y(1) = 8 y' 4xy = x^3 e^x^2 y(0) = 1 (x1) dy/dx y ln x y(1) = 10 x(x 1) dy/dx xy = 1 y(e) = 1 y' (sin x)y = 2 sin x y(/2) = 1 y' (tan x)y = cos^2 x y(0) = 1 In Problems 3740 proceed as in Example 6 to solve the given initialvalue problem Use a graphing utility to graph the continuous function y(x)1/2yx5/2=0;y2x3=0 No solution System of Linear Equations entered 1 1/2yx5/2=0 2 y2x3=0 Equations Simplified or Rearranged 1 y/2 x = 5/2 2 y 2x =
General Solution Of Y 4y 5y 2y 0 Youtube