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求微分方程的特解 y'-2y/(1-x^2)=x+1 x=0,y=0
人气:341 ℃ 时间:2020-05-09 05:25:43
解答
积分因子为exp(∫-2/(1-x^2 ) dx)=(x-1)/(x+1)微分方程两边同时乘(x-1)/(x+1),得(x-1)/(x+1)*y'+2*y/(x+1)^2=x-1即((x-1)/(x+1)*y)'=x-1两边积分并结合初始条件得(x-1)/(x+1)*y=1/2*x^2-x则y=1/2*x*(x-2)*(x+1)/(x-1...
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