高数题微分方程

Question

如题所述

Answer 1

（1）解：∵y'+y"=xy"
==>y'-(x-1)dy'/dx=0
==>y'dx-(x-1)dy'=0
==>dx/y'-(x-1)dy'/y'²=0 (等式两端同除y'²)
==>d((x-1)/y')=0
==>(x-1)/y'=C1 (C1是常数)
==>(x-1)/y'=1 (∵y'(2)=1，y(2)=1∴C1=1)
==>y'=x-1
==>y=x²/2-x+C2 (C2是常数)
==>y=x²/2-x+1 (∵y(2)=1∴C2=1)
∴所求特解是y=x²/2-x+1。
（2）解：∵y"-2yy'³=0
==>y'dy'/dy-2yy'³=0
==>dy'/dy-2yy'²=0 (等式两端同除y')
==>dy'/y'²-2ydy=0
==>1/y'+y²=C1 (C1是常数)
==>1/y'+y²=0 (∵y'(0)=-1，y(0)=1∴C1=0)
==>dx/dy+y²=0
==>dx+y²dy=0
==>x+y³/3=C2 (C2是常数)
==>x+y³/3=1/3 (∵y(0)=1∴C2=1/3)
==>3x+y³=1
∴所求特解是3x+y³=1。