根号y^2-xy dx-搜答案-搜搜考试网

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

（根号y/根号x-根号y)-(根号y/根号x+根号y)=｛根号y（根号x+根号y）｝/（x-y）-｛根号y（根号x-根号y）｝/(x-y)=(y+y)/(x-y)因为x=2y所以原式=2y/y=2

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

分离变量法xydx+√(1-x^2)dy=0dy/y=-xdx/√(1-x^2)dy/y=0.5d(1-x^2)/√(1-x^2)积分：ln|y|=√(1-x^2)+C1得:y=Ce^(√(1-x^2

：∵(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

答案在插图 :

原式=√y/(√2y-√y)-√y/(√2y+√y)=√y/[√y(√2-1)]-√y/[√y(√2+1)]=1/(√2-1)-1/(√2+1)=(√2+1)/(√2+1)(√2-1)-(√2-1)/

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

(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).

详细过程点下图查看

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