xydx-(x^2-y^2)dy=0
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本解答从这一步出发:得到∫[2(-y+t+1)y]d(-y+t+1)+[(-y+t+1)^2+f(y)]dy=0(y从t到1)也即∫[-2(-y+t+1)y]dy+[(-y+t+1)^2+f(y)]d
Q(x,y)=x^2+2y+1
∵(y²-3x²)dy+2xydx=0∴((y/x)²-3)dy+2(y/x)dx=0.(1)设t=y/x,则dy=xdt+tdx代入(1)得(t²-3)(xd
以D(X+Y)为例:D(X+Y)=E[(X+Y)-E(X+Y)]^2←方差的定义=E[X-E(X)+Y-E(Y)]^2=E[X-E(X)]^2+E[Y-E(Y)]^2+2E【[X-E(X)][Y-E(
dy/dx=2xy/(x^2-y^2)=(2y/x)/(1-(y/x)^2)令y/x=uy=ux,dy/dx=u+xdu/dx所以原式变为:u+xdu/dx=2u/(1-u^2)xdu/dx=(u+u
:∵(y²-3x²)dy+2xydx=0∴((y/x)²-3)dy+2(y/x)dx=0.(1)设t=y/x,则dy=xdt+tdx代入(1)得(t²-3)(x
第一题:原式左=(2xydx+x^2dy)+cosydy=d(x^2*y)+d(Siny)=d(X^2*y+Siny)=0所以通解为x^2*y+siny=C,C为常数第二问:变形为dy/dx=(y^2
∵(y^4-3x²)dy+xydx=0==>[(y^4-3x²)dy+xydx]/y^7=0==>dy/y³-3x²dy/y^7+xdx/y^6=0==>-d(
由T的参数方程及关于坐标的曲线积分公式得:原式=∫(0→π)[acost*asint*(-asint)+(acost-asint)*acost+(acost)^2*b]dt=a^2(1+b)π/2再问
这题我貌似再哪本书上看到过==||%2d是输入两列数,例如123456只会输入前面的两列数,12后面的数都被舍去了%*2d是跳过这个输入,也就是说,虽然那里有三个%d但实际上只读入了两个数而已如键盘输
z是[10~1/2]?如果是的话,答案是171/8;(可以把所求式子化为∫xdx*∫ydy*∫dz,再代入积分区间:原式=(2^2-1^2)/2*(1^2-(-2)^2)/2*(1/2-10)=171
(1+x^2)dy+2xydx=0(1+x^2)dy=-2xydx1/y*dy=-2x/(1+x^2)*dx两边同时积分得∫1/y*dy=∫-2x/(1+x^2)*dxln|y|=-ln|1+x^2|
(y^2-3x^2)/(2xy)=dx/dy,dx/dy=(y/x)/2-(3/2)(x/y),(1)设v=x/y,x=vy,dx/dy=1/(2v)-3v/2,(2)dx/dy=v+ydv/dy,(
xydx+(1+x^2)dy→(1/2)·[1/(1+x^2)]dx^2+(1/y)dy=0∴(1/2)ln(1+x^2)+lny+C=0.也可表为:y^2=C/(1+x^2).
设y/x=t,则y=xt,dy=xdt+tdx∵(y²+x²)dy-xydx=0==>(y/x+x/y)dy-dx=0==>(t+1/t)(xdt+tdx)=dx==>x(t
dy/dx=2xyy'/y=2x(lny)'=2xlny=x^2+Ay=e^(x^2+A)+B其中A,B是常数项
y²=x==>y=±√x∫_L(xy)dx=∫_(点A到原点)(xy)dx+∫_(原点到点B)(xy)dx=∫(1~0)x(-√x)dx+∫(0~1)x(√x)dx=∫(0~1)(x√x+x