x^2*y 3xy y=0的通解
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(1+x*x)=C(1+y*y)
(1)显然,y=0是原方程的解(2)若y≠0时,令y=xt,则dy=xdt+tdx代入原方程,化简得2dx/x=-dt/t^2==>2ln│x│=1/t+ln│C│(C是非零常数)==>x^2=Ce^
∵x²dy+(y-2xy-x²)dx=0==>e^(-1/x)dy/x²+(y-2xy-x²)e^(-1/x)dx/x^4=0(等式两端同乘e^(-1/x)/x
x(1+y²)dx=y(1+x²)dyx/(1+x²)dx=y/(1+y²)dy2x/(1+x²)dx=2y/(1+y²)dy1/(1+x&
∵dy/dx-2+y/x=0==>dy-2dx+ydx/x=0==>xdy-2xdx+ydx=0==>(xdy+ydx)-2xdx=0==>d(xy)-d(x^2)=0==>xy-x^2=C(C是常数
y'、dy/dx称为导数或微商.y'是dy/dx的简略写法,对默认自变量求导数.比如y=f(t),y'就是dy/dt.dy是微分,是差分的极限形式.dy=y'dx.严格地说,dy/dx不是dy与dx的
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∵(x-2y)dy+dx=0==>xe^ydy-2ye^ydy+e^ydx=0(等式两端同乘e^y)==>xd(e^y)+e^ydx=2yd(e^y)==>d(xe^y)=2yd(e^y)==>∫d(
∵dx+(x+y^2)dy=0==>e^ydx+xe^ydy+y^2e^ydy=0(等式两端同乘e^y)==>e^ydx+xd(e^y)+y^2e^ydy=0==>d(xe^y)+d((y^2-2y+
解法简单我们知道(y/x)'=(xy'-y)/x^2很容易就可以化简成(y/x)'=1所以解就是(y/x)'=x+C;把x乘过来就是y=x^2+Cx
1、dy=(2x+1)dx,y=x^2+x+C,2、dy/y=2dx,lny=2x+lnC1,y=e^(2x+lnC1),y=C*e^(2x).
一阶线性常系数,可以有两种方法第一种,设函数u=u(x),与原式子相乘,使得等式左边=d(uy)/dxuy'+2uy=uxe^x由乘法法则可得du/dx=2udu/u=2dx∫du/u=∫2dxu=e
y(x)=C1*BesselJ(v,x)+C2*BesselY(v,x)
对应的齐次方程为y'+2y=0解得y*=C(1)[e^(-2x)]然后用常数变易法求原方程的解,设原方程的解为y=C(x)[e^(-2x)]则y'=C'(x)[e^(-2x)]+C(x)(-2)[e^
即xy'-y=x^3即(xy'-x'y)/x^2=x即(y/x)'=xy/x=1/2x^2+cy=x(1/2x^2+c);c为常数
xy''+y'=x^2(xy')'=x^2xy'=1/3x^3+c1y'=1/3x^2+c1/xy=1/9x^3+c1ln[x]+c2
可分离变量型,原微分方程可化为dx/(1+x^2)=dy/(ylny),两边同时积分J1/(1+x^2)dx=J1/(lny)d(lny),得lnlny=arctanx+C1得通解lny=Ce^(ar