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Dy Dx yCotx 2xsinx 0

两边对x求导得5y^4y'+2y'-1-3x^6=0整理得y'=dy/dx=(3x^6+1)/(5y^4+2)令x=0则y=0从而dy/dx│x=0=0.5

y'-ycotx=0 dy/dx=y*cotx 1/y*dy=cotx*dx ln|y|=ln(sin(x))+c1 y=c2*sin(x) 令 y=c(x)*sin(x) 常数变易法 y'=c(x)cos(x)+c'(x)*sin(x) y'-y=c(x)cos(x)+c'(x)*sin(x)-c(x)cos(x)=2x*sin(x) => c'(x)=2x => c(x)=x^2 + c3 即y'-ycotx=2xsinx的通...

即∫x *2(sinx)^2dx =∫x *(1-cos2x)dx =∫x dx -∫ x *cos2x dx =0.5x^2 -∫0.5x d(sin2x) =0.5x^2 -0.5x *sin2x +∫0.5sin2x dx =0.5x^2 -0.5x *sin2x -0.25cos2x +C,C为常数

解:∵(ycosx+2xe^y)dx+(sinx+x^2e^y+2)dy=0 ==>(ycosxdx+sinxdy)+(2xe^ydx+x^2e^ydy)+2dy=0 ==>∫(ycosxdx+sinxdy)+∫(2xe^ydx+x^2e^ydy)+∫2dy=0 ==>ysinx+x^2e^y+2y=C (C是任意常数) ∴此方程的通解是ysinx+x^2e^y+2y=C。

上限是π/2,下限是0么 那么 ∫(2x+sinx)dx = x² -cosx (代入上下限π/2和0) =π²/4 +1

待续

补线用格林公式、满意请采纳,谢谢。

dy/dx=2-cosx>0 所以y=2x-sinx在R上为增函数,当然 y=2x-sinx在(0,兀)上为增函数

证明:由题意,P=2xcosy-y2sinx,Q=2ycosx-x2siny,在整个平面上具有一阶连续偏导数,且?P?y=?2xsiny?2ysinx=?Q?x∴曲线积分I与积分路径无关.取路径从(0,0)到(2,0)再到(0,3),则I=∫202xdx+∫30(2ycos2?4siny)dy=4+9cos2+4cos3-4=9cos...

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