(3)(y^2-3x^2)dy+2xydx=0,y|_(x=0)=1.

2）(y^2-3x^2)dy+2xydx=0；2）$(y^{2}-3x^{2})dy+2xydx=0$；

2.求下列齐次方程满足所给初值条件的特解：(1)(y^2-3x^2)dy+2xydx=0,y|_(x=0)=1；2.求下列齐次方程满足所给初值条件的特解：$(1

2.求下列齐次方程满足所给初值条件的特解:-|||-(1) ((y)^2-3(x)^2)dy+2xydx=0 |x=0=1;-|||-(2) =dfrac (x

微分方程^3dx+(2x(y)^2-1)dy=0的通解为 A ^3dx+(2x(y)^2-1)dy=0（^3dx+(2x(y)^2-1)dy=0 为任意常数）B

(3)(x^2+2xy-y^2)dx+(y^2+2xy-x^2)dy=0,y|_(x=1)=1;(3)$(x^{2}+2xy-y^{2})dx+(y^{2}+2

(2)(x^2+2xy-y^2)dx+(y^2+2xy-x^2)dy=0, y|_(x=1)=1.(2)$(x^{2}+2xy-y^{2})dx+(y^{2}+

1.微分方程(1-(x)^2)(y)^2dfrac (dy)(dx)+(2(x)^2-1)(y)^3=(x)^3是(1-(x)^2)(y)^2dfrac (dy

微分方程((y)^2+2)dx+y((x)^2+1)dy=0的通解为( )((y)^2+2)dx+y((x)^2+1)dy=0((y)^2+2

1.计算:(3x+y+2)(3x-y-2).2.已知(3x+y+2)(3x-y-2).求（a+b-c）²。3.设x>0,且(3x+y+2)(3x-y-2).求(

(4)(dy)/(dx)+3y=8,y|_(x=0)=2;(4)$\frac{dy}{dx}+3y=8,y|_{x=0}=2;$