微积分题dy/dx=(-2x+y)^2-7,y(0)=0求y

Question

微积分题
dy/dx=(-2x+y)^2-7,y(0)=0 求y

Answer 1

设z=-2x+y,则dy/DX=dz/DX+2
代入原方程得dz/DX+2=z²-7
==>dz/DX=z²-9
==>dz/(z²-9)=dx
==>[1/(z-3)-1/(z+3)]dz=6dx
==>ln│z-3│-ln│z+3│=6x+ln│C│ (C是积分常数)
==>ln│(z-3)/(z+3)│=6x+ln│C│
==>(z-3)/(z+3)=Ce^(6x)
==>(y-2x-3)/(y-2x+3)=Ce^(6x)
∴原微分方程的通解是(y-2x-3)/(y-2x+3)=Ce^(6x)
∵y(0)=0
∴-3/3=C ==>C=-1
故满足初始条件的特解是(y-2x-3)/(y-2x+3)=-e^(6x).