1.dy/dx=(xy^2-cosxsinx)/(y(1-x^2)),y(0)=2求y2.xydx+(2x^2+3y^2-20)dy=0,y(0)=1求y3.dy/dx=(-2x+y)^2-7,y(0)=0求y

Question

1.dy/dx=(xy^2-cosxsinx)/(y(1-x^2)),y(0)=2 求y
2.xydx+(2x^2+3y^2-20)dy=0,y(0)=1 求y
3.dy/dx=(-2x+y)^2-7,y(0)=0 求y

Answer 1

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).