（x＋1）y＇-2y-（x＋1）＝0y＇-2y/（x＋1）＝1一般情况下：y'+p(x)y=q(x)那么其解的公式为：y=e^[-∫p(x)dx]{∫q(x)*e^[∫p(x)dx]dx+C}此处p(x)=-2/(x+1),q(x)=1y=e^[-∫p(x)dx]{∫q(x)*e^[∫p(x)dx]dx+C}=e^[∫2/(...