Question

微积分题目

Answer 1

解：1。∵dy/dx=(xy2-cosxsinx)/(y(1-x2)) ==>y(1-x2)dy=(xy2-cosxsinx)dx ==>y(1-x2)dy-xy2dx+cosxsinxdx=0 ==>(1-x2)d(y2)-y2d(x2)+sin(2x)dx=0 ==>2(1-x2)d(y2)+2y2d(1-x2)+sin(2x)d(2x)=0 ==>2d(y2(1-x2))+sin(2x)d(2x)=0 ==>2y2(1-x2)-cos(2x)=C (C是积分常数) ∴原微分方程的通解是2y2(1-x2)-cos(2x)=C (C是积分常数) ∵ y(0)=2 ∴8-1=C ==>C=7 故满足初始条件的特解是2y2(1-x2)-cos(2x)=7； 2。∵xydx+(2x^2+3y^2-20)dy=0 ==>xy^4dx+2x2y^3dy+3y^5dy-20y3dy=0 (等式两边同乘y^3) ==>y^4d(x2)/2+x2d(y^4)/2+d(y^6)/2-5d(y^4)=0 ==>d(x2y^4)+d(y^6)-10d(y^4)=0 ∴原微分方程的通解是x2y^4+y^6-10y^4=C (C是积分常数) ∵y(0)=1 ∴1-10=C ==>C=-9 故满足初始条件的特解是x2y^4+y^6-10y^4=-9； 3。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)。