求(1+x)y''+y'=(1+x)^2满足初始条件y(0)=1,y'(0)=-1的特解
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求(1+x)y''+y'=(1+x)^2满足初始条件y(0)=1,y'(0)=-1的特解
求(1+x)y''+y'=(1+x)^2满足初始条件y(0)=1,y'(0)=-1的特解
求(1+x)y''+y'=(1+x)^2满足初始条件y(0)=1,y'(0)=-1的特解
解答见附图:
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