2.求下列微分方程的通解:(1)2xydy-(x^2+2y^2)dx=0 ;(2 (dy)/(dx)=y/x+tany/x ;(3) (x^3+y^3)dx-3xy^2d
2.求下列微分方程的通解:(1) 2xydy-(x^2+2y^2)dx=0 ;(2) (dy)/(dx)=y/x+tany/x ;(3) (x^3+y^3)dx-3xy^2dy=0 ;(4)(y+x)dy+(y-x)dx=0;(5) x(dy)/(dx)=y+√(x^2-y^2) :(6) x(lnx-lny)dy-ydx=0 ;(7) (3x^2+2xy-y^2)dx+(x^2-2xy)d...
dy/dx=(2x^3+3xy^2+x)/(3x^2y+2y^3-y),右边分母分子分别提取公因式x，y，则：dy/dx=x(2x^2+3y^2+1)/y(3x^2+2y^2-1),将右边提出的x，y移动到等号左边。ydy/xdx=(2x^2+3y^2+1)/(3x^2+2y^2-1),左边凑分分别到dy、dx中，得：dy^2/dx^2=(2x^2+3y^2+1)/(3x^2+2y^2-1)...
ydy/xdx=(2x^2+3y^2+1)/(3x^2+2y^2-1),dy^2/dx^2=(2x^2+3y^2+1)/(3x^2+2y^2-1)。dy^2/dx^2=[2(x^2-1)+3(y^2+1)]/[3(x^2-1)+2(y^2+1)]. 设：u=(y^2+1)/(x^2-1),则：u(x^2-1)=y^2+1,两边求全微分得：(x^2-1)du+udx^2=dy^2 ...
,又点Q在原曲线\Gamma : {x}^{2} + {y}^{2} - {xy} = 1上,所以{\left( \frac{x - y}{\sqrt{2}}\right) }^{2}+ {\left( \frac{x + y}{\sqrt{2}}\right) }^{2} - \frac{x - y}{\sqrt{2}} \cdot \frac{x + y}{\sqrt{2}} = 1,整理得:3{x}^{2} + {y}...
解(1)原方程可化为 y'-2/xy=x^2cosx .这是一阶线性非齐 次方程.首先考虑它所对应的一阶线性齐次方程 y'-2/xy=0 , 可得其通解y=Cx2.利用“常数变易法” ,令原方程的通解为 y=u(x)x2, 则 y'=u'x2+2ux. 代入原微分方程,得 u'x2=x2cosx 或 u'=cosx, 于是 u t=∫cosx...
D(XY) = D(X)D(Y)解题过程如下：D(XY) = E{[XY-E(XY)]^2} = E{X²Y²-2XYE(XY)+E²(XY)} = E(X²)E(Y²)-2E²(X)E²(Y)+E²(X)E²(Y)= E(X²)E(Y²)-E²(X)E²(Y)如果 E(X) ...
∫∫(2x+3y)dxdy=∫∫(3x+2y)dxdy =2.5∫(∫x+y)dxdy =2.5 ∫ [0,2]dx ∫ [0,2-x](x+y)dy =2.5 ∫ [0,2][x(2-x)+0.5(x-2)^2]dx =2.5∫ [0,2][1-0.5x^2]dx =5/3 分析总结。 计算二重积分d2x3ydxdy其中d是由两坐标轴及直线xy2所围成的闭区域结果...
Q = 3x²y + 2y²、Q'x = 6xy 所以是恰当方程。令原函数为u(x，y)∂u/∂x = x² + 3xy²u = ∫ (x² + 3xy²) dx = x³/3 + 3x²y²/2 + φ(y)∂u/∂y = 3x²y + φ'(y) = 3x...
x^2y+3xy=sin2x解微分方程两边同时对x求导得2xy+x²y’+3y+3xy’=2cos2xy’（x²+3x）=2cos2x-2xy-3yy’=（2cos2x-2xy-3y）/（x²+3x）所以dy=（2cos2x-2xy-3y）/（x²+3x）dx x^2y'+3xy=sin2x解微分方程 您这边不是求微分方程吗？您这边没有说求通解 解微分方程老师...