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(
已知函数 y = y(x) 满足方程 xydx = sqrt(2 - x^2) dy ,且当 x = 1 时 y = 1 ,则当 x = -1 , y = (