求微分方程的特解y'-2y/(1-x^2)=x+1 x=0,y=0

求微分方程的特解y'-2y/(1-x^2)=x+1 x=0,y=0

積分因數為exp(∫-2/(1-x^2)dx)=(x-1)/(x+1)微分方程兩邊同時乘(x-1)/(x+1),得(x-1)/(x+1)*y'+2*y/(x+1)^2=x-1即((x-1)/(x+1)*y)'=x-1兩邊積分並結合初始條件得(x-1)/(x+1)*y=1/2 *x^2-x則y=1/2*x*(x-2)*(x+1)/(x-1…