Question

求微分方程y'-y/x=xe^x的通解

Answer 1

-1/x积分得:-lnx+C1

令f(x)=e^(lnx+C1),则f'(x)=e^(-lnx+C)/(-x)=f(x)/(-x)

原式两边乘以f(x)

f(x)y'-yf(x)/x=xe^x*f(x)

f(x)y'+yf'(x)=x*e^x*e^(lnx+C1)=x*e^x*(x+e^C1)=x^2*e^x+x*e^x*e^C1

两边积分:

f(x)*y=x^2*e^x+(e^c1-2)x*e^x+(2-e^c1)*e^x+C2

y=e^(lnx+C1)*[x^2*e^x+(e^c1-2)x*e^x+(2-e^c1)*e^x+C2]