微积分题目 微积分题目？
\u5fae\u79ef\u5206\u9898\u76ee\u89e3\uff1a1\u3002\u2235dy/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\u662f\u79ef\u5206\u5e38\u6570) \u2234\u539f\u5fae\u5206\u65b9\u7a0b\u7684\u901a\u89e3\u662f2y2(1-x2)-cos(2x)=C (C\u662f\u79ef\u5206\u5e38\u6570) \u2235 y(0)=2 \u22348-1=C ==>C=7 \u6545\u6ee1\u8db3\u521d\u59cb\u6761\u4ef6\u7684\u7279\u89e3\u662f2y2(1-x2)-cos(2x)=7\uff1b 2\u3002\u2235xydx+(2x^2+3y^2-20)dy=0 ==>xy^4dx+2x2y^3dy+3y^5dy-20y3dy=0 (\u7b49\u5f0f\u4e24\u8fb9\u540c\u4e58y^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 \u2234\u539f\u5fae\u5206\u65b9\u7a0b\u7684\u901a\u89e3\u662fx2y^4+y^6-10y^4=C (C\u662f\u79ef\u5206\u5e38\u6570) \u2235y(0)=1 \u22341-10=C ==>C=-9 \u6545\u6ee1\u8db3\u521d\u59cb\u6761\u4ef6\u7684\u7279\u89e3\u662fx2y^4+y^6-10y^4=-9\uff1b 3\u3002\u8bbez=-2x+y\uff0c\u5219dy/dx=dz/dx+2 \u4ee3\u5165\u539f\u65b9\u7a0b\u5f97dz/dx+2=z2-7 ==>dz/dx=z2-9 ==>dz/(z2-9)=dx ==>[1/(z-3)-1/(z+3)]dz=6dx ==>ln\u2502z-3\u2502-ln\u2502z+3\u2502=6x+ln\u2502C\u2502 (C\u662f\u79ef\u5206\u5e38\u6570) ==>ln\u2502(z-3)/(z+3)\u2502=6x+ln\u2502C\u2502 ==>(z-3)/(z+3)=Ce^(6x) ==>(y-2x-3)/(y-2x+3)=Ce^(6x) \u2234\u539f\u5fae\u5206\u65b9\u7a0b\u7684\u901a\u89e3\u662f(y-2x-3)/(y-2x+3)=Ce^(6x) \u2235y(0)=0 \u2234-3/3=C ==>C=-1 \u6545\u6ee1\u8db3\u521d\u59cb\u6761\u4ef6\u7684\u7279\u89e3\u662f(y-2x-3)/(y-2x+3)=-e^(6x)\u3002
\u8fd9\u9898\u662f\u6709\u4e24\u4e2a\u683c\u62c9\u6717\u65e5\u4e2d\u503c\u5b9a\u7406\u7684\uff0c\u5de6\u53f3\u540c\u9664\u4ee5(b-a)\uff0c\u5de6\u8fb9\u53ef\u4ee5\u5f97\u5230[f(b)-f(a)]/(b-a)\uff0c\u8fd9\u662ff(x)\u7684\u4e2d\u503c\u5b9a\u7406\uff0c\u53f3\u8fb9\u53ef\u4ee5\u5f97\u5230(b^2-a^2)/(b-a)\uff0c\u8fd9\u662fx^2\u7684\u4e2d\u503c\u5b9a\u7406\uff0c\u5b83\u7684\u5bfc\u6570\u6b63\u597d\u662f2x\uff0c\u5f97\u6b63\u597dx^2\u548cf(x)\u7684\u62c9\u683c\u6717\u65e5\u70b9\u4e00\u6837\uff0c\u624d\u80fd\u5f97\u5230\u65b9\u7a0b\u7684\u89e3\uff0c\u800c\u6761\u4ef6\u4e2d\u6839\u672c\u65e0\u6cd5\u786e\u5b9a\u5b83\u4eec\u7684\u62c9\u683c\u6717\u65e5\u70b9\u4e00\u81f4\uff0c\u6240\u4ee5\u662f\u9519\u9898\u3002
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=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)
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)。
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