Question

求微分方程+y'+2xy=4x+在初始条件+y|_(x=0)=3+下的特解

Answer 1

问：求微分方程+y'+2xy=4x+在初始条件+y|_(x=0)=3+下的特解

答：y' + 2xy = 4x dy/dx = 2x(2 - y) dy/(2 - y) = 2x dx -ln(2 - y) = x² + C ln(2 - y) = -x² + C 2 - y = e^(-x² + C) -y = Ce^(-x²) - 2 y = Ce^(-x²) + 2 第二题： y' - ytanx = secx e^∫ -tanx dx = e^-(-lncosx) = cosx cosx * y' - cosx * ytanx = secx * cosx cosx * y' - ysinx = 1 (ycosx)' = 1 ycosx = x + C y = xsecx + Csecx y|(x=0) = 0 得 C = 0 ∴y = xsecx y' + y/x = sinx/x e^∫ 1/x dx = e^lnx = x x * y' + y = sinx (xy)' = sinx xy =- cosx + C y = (C - cosx)/x y|(x=π) = 1 1 = (C - cosπ)/π C = π + 1 ∴y = (-cosx)/x +