解方程y´=x(2y+1)/(1+2)
y′=-[x(2y+1)/(1+x2)](可分离变量)[1/(2y+1)]dy=-[x/(1+x2dx)]⇒(1/2)ln∣2y+1∣=-(1/2ln)(1+x2)+lnC1/2⇒2y+1=C/(1+2)(C=±C1),即(1+x2)(2y+1)=C
解方程y´=x(2y+1)/(1+2)
y′=-[x(2y+1)/(1+x2)](可分离变量)[1/(2y+1)]dy=-[x/(1+x2dx)]⇒(1/2)ln∣2y+1∣=-(1/2ln)(1+x2)+lnC1/2⇒2y+1=C/(1+2)(C=±C1),即(1+x2)(2y+1)=C