Question

求微分方程(x+2)y'-(x^2)y=0

Answer 1

变量分离,y'(1/y)=(x^2)/(x+2),(1/y)dy=dx(x^2)/(x+2),两边同时积分,注意dx(x^2)/(x+2)可化成dx(x^2-4)/(x+2)+4dx/(x+2) => (x-2)dx+4d(x+2)/(x+2) 再求积分,得y=exp(0.5x^2-2x+4ln(x+2))+c=exp(0.5x^2-2x)+(x+2)^4+c