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y=Cos1/x,求Dy/Dx

解:1/(1+y^2)y'=1/(1-(1/x)^2)^1/2*(-1/x^2) y'=(1+y^2)/x(x^2-1)^1/2。 答:y'=(1+y^2)/x(x^2-1)^1/2。

利用隐函数定理. 记f(x,y)=cos(xy)-ln((x+1)/y)-1,则函数y=y(x)由f(x,y)=0确定,则dy/dx=-(df/dx)/(df/dy)=-(-sin(xy)*y-y/(x+1)/y)/(-sin(xy)*x-y/(x+1)*(x+1)/(-y^2))=(sin(xy)*y+1/(x+1))/(-sin(xy)*x+1/y),把x=0代入方程,解得y=1,代入dy...

dcos(xy)+dlny-dx=d(1) -sin(xy)d(xy)+(1/y)dy-1=0 -sin(xy)(xdy+ydx)+(1/y)dy-1=0 -xsin(xy)dy-ysin(xy)dx+(1/y)dy-1=0 所以dy/dx=[ysin(xy)+1]/[1/y-xsin(xy)] =[y²sin(xy)+y]/[1-xysin(xy)]

不能作答

dy/dx=-(x*x)+cosx+1/x+e^x 都是积分表里需要背的,记住就行,没步骤啊

求导,y'=ysinx+x*(y'sinx+y*cosx)-sin(x-1) 移项 y'-xy'sinx=ysinx+xycosx-sin(x-1) y'=(ysinx+xycosx-sin(x-1))/(1-xsinx)

解:∵[1+ycos(xy)]dx+xcos(xy)dy=0 ==>dx+ycos(xy)dx+xcos(xy)dy=0 ==>dx+cos(xy)(ydx+xdy)=0 ==>dx+cos(xy)d(xy)=0 ==>∫dx+∫cos(xy)d(xy)=0 ==>x+sin(xy)=C (C是积分常数) ∴此方程的通解是x+sin(xy)=C。

先纠正你的两题答案的打印错误:第一题的答案应该是y=C1+C2*e^x-x; 第二题的答案应该是y=-ln|cos(x+C1)|+C2 说明:这两题不要用p代y’,这样反而搞复杂了。甚至求不出方程的解。 正确解法如下: 解:1.∵y''=1+y' ==>dy'/dx=1+y' ==>dy'/(1+y')=dx...

不详

本题主要求y=x²的极坐标方程,即rsinθ=r²cos²θ,整理后为:r=sinθ/cos²θ则∫(0->1)dx∫(x^2->x)(x^2+y^2)^(-1/2)dy=∫[0->π/4]dθ∫[0->sinθ/cos²θ] (1/r)*rdr=∫[0->π/4]dθ∫[0->s...

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