ydx-xdy的积分因子

汲中奇4995(1+x*2)ydx - (2 - y)xdy=0求通解 -
张胥琰19442173752 ______ 解:∵(1+x^2)ydx-(2-y)xdy=0 ==>(1+x^2)ydx=(2-y)xdy ==>(2-y)dy/y=(1+x^2)dx/x ==>(2/y-1)dy=(1/x+x)dx ==>2ln│y│-y=ln│x│+x^2/2+ln│C│ (C是常数) ==>y^2*e^(-y)=Cxe^(x^2/2) ==>y^2=Cxe^(y+x^2/2) ∴原方程的通解是y^2=Cxe^(y+x^2/2).

汲中奇4995(x+y)dy–ydx=0通解 -
张胥琰19442173752 ______ 解:∵(x+y)dy–ydx=0 ==>ydy-(ydx-xdy)=0 ==>dy/y-(ydx-xdy)/y^2=0 (等式两端同除y^2) ==>dy/y-d(x/y)=0 ==>ln│y│-x/y=ln│C│ (C是常数) ==>ye^(-x/y)=C ==>y=Ce^(x/y) ∴原方程的通解是y=Ce^(x/y).

汲中奇4995解下列微分方程:ydx - xdy=ysec(x/y)dy,ylx=0=1 -
张胥琰19442173752 ______ cos(x/y)[(ydx-xdy)/y^2]=(1/y)dy cos(x/y)d(x/y)=d(lny) sin(x/y)=lny+c 初始条件代人得:0=0+c c=0 ∴所求微分方程初值问题的解为 sin(x/y)=lny

汲中奇4995微分方程yln ydx+dy=0的通解怎么求 -
张胥琰19442173752 ______ 思路:首先移项,使不同的变量分别在等号的两边,然后两边同时积分,即可求出答案.(此方法叫作分离变量法) 详细解答如下:∵y Iny dx+ dy=0 ∴dx=-1/[yIny]·dy ∴x+C1=-In(Iny) ∴C·e^x=1/Iny 即C·e^x·Iny=1

汲中奇4995解下列微分方程:ydx - xdy=ysec(x/y)dy,ylx=0=1
张胥琰19442173752 ______ 令x/y=u,则x=yu 则dx=udy+ydu 代入可得: yudy+y^2du-yudy=y*secu*dy 即cosudu=dy/y 积分得: ln|y|=sinu+A y=±e^A*e^{sinu}=B*e^{sin(x/y)} 由x=0时,y=1得: 1=B 从而得特解为: y=e^{sin(x/y)}

汲中奇4995ydx - (1+x+y^2)dy =0通解 -
张胥琰19442173752 ______ 解:∵ydx-(1+x+y^2)dy=0 ==>(ydx-xdy)-(1+y^2)dy=0 ==>(ydx-xdy)/y^2-(1/y^2+1)dy=0 (等式两端同除y^2) ==>d(x/y)-(1/y^2+1)dy=0 ==>∫d(x/y)-∫(1/y^2+1)dy=0 ==>x/y+1/y-y=C (C是积分常数) ==>x=y^2+Cy-1 ∴原方程的通解是x=y^2+Cy-1.

汲中奇4995微分方程 (y - x)y'+y=0满足y(0)=1的特解为 -
张胥琰19442173752 ______[答案] (y-x)y'+y=0 ydy-xdy+ydx=0 ydy+ydx-xdy=0 dy/y+[ydx-xdy]/y^2=0 dy/y+d(x/y)=0 积分得:lny+(x/y)=lnC y=Ce^(-x/y) y(0)=1代入:C=1 y=e^(-x/y)

汲中奇4995l用积分因子法求方程(y - x*2)dx - xdy=0的通解 -
张胥琰19442173752 ______[答案] ∵(y-x^2)dx-xdy=0 ==>xdy-ydx+x^2dx=0 ==>(xdy-ydx)/x^2+dx=0 (等式两端同乘积分因子1/x^2) ==>d(y/x)+dx=0 ==>∫d(y/x)+∫dx=0 ==>y/x+x...

汲中奇4995用Matlab求解常微分方程求初值问题,原题是xdy+(x2–y)dx=0. -
张胥琰19442173752 ______ 先变形为dy/dx=y/x-x,再用dsolve求通解或ode45求数值解.如:syms y(x) y=dsolve(diff(y)==y/x-x) 结果是:y = - x^2 + C1*x

汲中奇4995计算(ydx - xdy)/(x2+y2),其中l为圆周(x - 1)2+(y - 2)2=4(逆时针) -
张胥琰19442173752 ______[答案] L: (x-1)^2+(y-2)^2=4, 原点(0, 0)在圆外, 故积分函数在圆内满足格林公式条件. P'=(x^2-y^2)/(x^2+y^2)^2, Q' = - (y^2-x^2)/(x^2+y^2)^2=P' ∮(ydx-xdy)/(x^2+y^2) = 0.