微积分题目

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(0)=2 求y
解：ydy/dx=(xy²-cosxsinx)/(1-x²)=xy²/(1-x²)-cosxsinx/(1-x²).............(1)
为了求(1)的解，可先考虑方程：ydy/dx=xy²/(1-x²)，消去y得 dy/dx=xy/(1-x²)，
分离变量得dy/y=xdx/(1-x²)=-d(1-x²)/[2(1-x²)]；
积分之得lny=-(1/2)ln(1-x²)+lnC₁=ln[C₁/√(1-x²)]
故得y=C₁/√(1-x²)..............(2)
把(2)中的任意常数C ₁换成x的函数u，于是y=u/√(1-x²)............(3)
对x取导数得：dy/dx=[(du/dx)/√(1-x²)]+[ux/√(1-x²)³]....................(4)
将(3)和(4)代入(1)式得：[u/√(1-x²)]{[(du/dx)/√(1-x²)]+[ux/√(1-x²)³]}=[xu²/(1-x²)²]-cosxsinx/(1-x²)
即有u(du/dx)/(1-x²)+xu²/(1-x²)²=xu²/(1-x²)²-cosxsinx/(1-x²)
于是得udu/dx=-cosxsinx，分离变量得udu=-cosxsinxdx=cosxd(cosx)
积分之得u²/2=(cos²x)/2+C/2，故u=cosx+C，再代入(3)即得通解y=(cosx+C)/√(1-x²),
将初始条件y(0)=2得2=1+C，故C=1，于是得特解为：y=(cosx+1)/√(1-x²).
2.xydx+(2x²+3y²-20)dy=0, y(0)=1 求y
解：将原式两边同乘以积分因子y³ï¼Œå¾—;
xy⁴dx+(2x²y³ +3y^5-20y³)dy=0............(1)
由于∂P/∂y=4xy³=∂Q/∂x，故(1)是全微分方程，于是得通解为：
[0，x]∫xy⁴dx+[0，y]∫(2x²y³ +3y^5-20y³)dy=(x²y⁴/2)+(x²y⁴/2)+(y^6)/2-5y⁴=C
即有x²y⁴+(y^6)/2-5y⁴=C
将初始条件x=0，y=1代入得C=1/2-5=-9/2
故得满足初始条件的特解为x²y⁴+(y^6)/2-5y⁴+9/2=0
去掉分母得2x²y⁴+y^6-10y⁴+9=0
3.dy/dx=(-2x+y)²-7, y(0)=0 求y
解：令u=-2x+y，则y=u+2x，故dy/dx=(dy/du)(du/dx)+d(2x)/dx=du/dx+2
于是有du/dx+2=u²-7，du/dx=u²-9，du/(u²-9)=(1/6)[1/(u-3)-1/(u+3)]du=dx，
积分之得(1/6)[ln(u-3)/(u+3)]=x+lnC，ln[(u-3)/(u+3)]=6x+lnC
将u=-2x+y代入即得通解：ln[(y-2x-3)/(y-2x+3)]=6x+lnC，即(y-2x-3)/(y-2x+3)=Ce^(6x)
将初始条件x=0，y=0代入得C=-1，故满足初始条件的特解为：
(y-2x-3)/(y-2x+3)=-e^(6x)

Answer 2

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)。本回答被提问者采纳