初值问题 =sqrt {{x)^2-(x)^2(y)^2+1-(y)^2} y(0)=-dfrac (1)(2) .

求下列初值问题的解：(dfrac (1+{x)^2}(2))y=arctan x，(dfrac (1+{x)^2}(2))y=arctan x.（ ）求下列初值

(B) dfrac (x)(y)((y+1))^2-|||-(C) ^2((x+dfrac {1)(x))}^2. (D) dfrac (y)(x)((y+1)

设=dfrac (2x)({x)^2-(y)^2} ，则 =dfrac (2x)({x)^2-(y)^2}=dfrac (2x)({x)^2-(y)^2}___

1.求下列齐次方程的通解:-|||-(1) -y-sqrt ({y)^2-(x)^2}=0;-|||-(2) dfrac (dy)(dx)=ycdot dfra

求下列函数的自然定义域：(1)y=sqrt(3x+2)；(2)y=dfrac(1)(1-{x)^2}；(3)y=dfrac(1)(x)-sqrt(1-(x)^2

1.求下列齐次方程的通解:-|||-(1) -y-sqrt ({y)^2-(x)^2}=0;-|||-(2) dfrac (dy)(dx)=yln dfrac

1.求下列齐次方程的通解:-|||-(1) -y-sqrt ({y)^2-(x)^2}=0;-|||-(2) dfrac (dy)(dx)=yln dfrac

证明:函数-|||-f(x,y)= ((x)^2+(y)^2)sin dfrac (1)(sqrt {{x)^2+(y)^2}}, ^2+(y)^2neq 0,

设函数 (x,y)=1-dfrac (cos sqrt {{x)^2+(y)^2}}(tan ({x)^2+(y)^2)} ,则当定设函数 (x,y)=1-df

-y-sqrt ({y)^2-(x)^2}=0;;