微积分考试题

(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).(2x²+3y²-20)dy=0, y(0)=1 求y解：将原式两边同乘以积分因子y³，得;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=(-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)