Question

高等数学求通解

Answer 1

1。8xy'-12y=-(5x²+3)y³,y(1)=√2
解：令z=1/y²，则y'=-y³z'/2
代入原方程化简整理，得
4xz'+12z=5x²+3..........(1)
==>4xdz+12zdx=(5x²+3)dx
==>4x³dz+12x²zdx=(5x^4+3x²)dx (等式两端同乘x²)
==>4x³dz+4zd(x³)=d(x^5+x³)
==>d(4x³z)=d(x^5+x³)
==>4x³z=x^5+x³+C (C是积分常数)
==>4x³/y²=x^5+x³+C
==>4x³=(x^5+x³+C)y²
∵y(1)=√2，则C=0
∴所求解是4x³=(x^5+x³)y²。
2。2(cos²ycos(2y)-x)y'=sin(2y),y(3/2)=5π/4
解：由2(cos²ycos(2y)-x)y'=sin(2y)
==>2(cos(2y)-x/cos²y)y'=sin(2y)/cos²y (等式两端同除cos²y)
==>2cos(2y)dy-2xdy/cos²y=2sinycosydx/cos²y
==>d(sin(2y))=2xd(tany)+2tanydx
==>d(sin(2y))=d(2xtany)
==>sin(2y)=2xtany+C (C是积分常数)
∵y(3/2)=5π/4，则C=-2
∴所求解是sin(2y)=2xtany-2。