y+ln+x+1
@籍重6059：求(x+1)y''+y'=ln(x+1)的通解? -
奚怕14796509783…… 具体回答如下: 令x+1=e^t,则(x+1)y'=dy/dt,(x+1)²y''=d²y/dt²-dy/dt. 代入原方程: dy/dt-dy/dt+dy/dt=te^t d²y/dt²=te^t dy/dt=∫te^tdt=te^t-e^t+C1(C1是积分常数) y=∫[(te^t-e^t+C1]dt =te^t-2e^t+C1t+C2(C2是积分常数) =(x+1)ln(x+...
@籍重6059：Y=1+ln(X+1)的反函数怎么算的?帮帮忙. -
奚怕14796509783…… Y=1+ln(X+1) y-1=ln(x+1) x+1=e^(y-1) x=e^(y-1)-1 交换字母顺序得 y=e^(x-1)-1
@籍重6059：证明函数y=2+ln(x+1)在( - 1,1]无界 -
奚怕14796509783…… x→-1+时x+1→0+,ln(x+1)→-∞, ∴函数y=2+ln(x+1)在(-1,1]无界.
@籍重6059：求解 [y+ln(x+1)]dx+(x+1 - e^y)dy是u(x,y)的全微分,则U(x,y)= -
奚怕14796509783…… 解;[y+ln(x+1)]dx+(x+1-e^y)dyydx+xdy+ln(x+1)dx+dy-e^ydy=d(xy)+ln(x+1)d(x+1)+dy-d(e^y)=d(xy)+d[(x+1)ln(x+1)-x]+dy-d(e^y)=d(xy+(x+1)ln(x+1)-x+y-e^y)u(x,y)=xy+(x+1)ln(x+1)-x+y-e^y
@籍重6059：y=x+ln(x+1)的导数的定义域是多少……? -
奚怕14796509783…… y=x+ln(x+1) y'=1+1/(x+1) x+1≠0 定义域:x≠-1
@籍重6059：求方程(x+1)y''+y'=ln(x+1)的通解时,若令y'=p,则: - 作业帮
奚怕14796509783…… [选项] A. y''=p' B. y''=p(dp)/(dy) C. y''=p(dp)/(dx) D. y''=p'(dp)/(dy)中那个选项正确,为什么?
@籍重6059：函数z=1/ln(x+y+1)的定义域 -
奚怕14796509783…… ln(x+y+1)≠0 【它充当分式的分母,当然不能为0】也就是ln(x+y+1)≠0 =ln1 x+y+1≠1且x+y+1>0【对数的真数必须大于0】联合得到:x+y∈(-1,0)∪(0,+∞)
@籍重6059：Y=(X+1)LN(X+1)求导?要过程. -
奚怕14796509783…… 乘法法则 y=(x+1) ln(x+1) y'=(x+1)' ln(x+1)+(x+1)[ln(x+1)]' =ln(x+1)+1
@籍重6059：求(x+1)y''+y'=ln(x+1)的通解? -
奚怕14796509783…… 解:令x+1=e^t,则(x+1)y'=dy/dt,(x+1)²y''=d²y/dt²-dy/dt 代入原方程得d²y/dt²-dy/dt+dy/dt=te^t ==>d²y/dt²=te^t ==>dy/dt=∫te^tdt=te^t-e^t+C1 (C1是积分常数) 于是,y=∫[(te^t-e^t+C1]dt =te^t-2e^t+C1t+C2 (C2是积分常数) =(x+1)ln(x+1)-2(x+1)+C1ln(x+1)+C2 故原方程的通解是y=(x+1)ln(x+1)-2(x+1)+C1ln(x+1)+C2 (C1,C2是积分常数)
@籍重6059：xy=ln(x+y)求导数步鄹 -
奚怕14796509783…… 两边对x求导,得 y+xy'=1/(x+y) ·(1+y') (x-1/(x+y))y'=1/(x+y)-y 所以 y'=【1/(x+y)-y】/(x-1/(x+y))
奚怕14796509783…… 具体回答如下: 令x+1=e^t,则(x+1)y'=dy/dt,(x+1)²y''=d²y/dt²-dy/dt. 代入原方程: dy/dt-dy/dt+dy/dt=te^t d²y/dt²=te^t dy/dt=∫te^tdt=te^t-e^t+C1(C1是积分常数) y=∫[(te^t-e^t+C1]dt =te^t-2e^t+C1t+C2(C2是积分常数) =(x+1)ln(x+...
@籍重6059：Y=1+ln(X+1)的反函数怎么算的?帮帮忙. -
奚怕14796509783…… Y=1+ln(X+1) y-1=ln(x+1) x+1=e^(y-1) x=e^(y-1)-1 交换字母顺序得 y=e^(x-1)-1
@籍重6059：证明函数y=2+ln(x+1)在( - 1,1]无界 -
奚怕14796509783…… x→-1+时x+1→0+,ln(x+1)→-∞, ∴函数y=2+ln(x+1)在(-1,1]无界.
@籍重6059：求解 [y+ln(x+1)]dx+(x+1 - e^y)dy是u(x,y)的全微分,则U(x,y)= -
奚怕14796509783…… 解;[y+ln(x+1)]dx+(x+1-e^y)dyydx+xdy+ln(x+1)dx+dy-e^ydy=d(xy)+ln(x+1)d(x+1)+dy-d(e^y)=d(xy)+d[(x+1)ln(x+1)-x]+dy-d(e^y)=d(xy+(x+1)ln(x+1)-x+y-e^y)u(x,y)=xy+(x+1)ln(x+1)-x+y-e^y
@籍重6059：y=x+ln(x+1)的导数的定义域是多少……? -
奚怕14796509783…… y=x+ln(x+1) y'=1+1/(x+1) x+1≠0 定义域:x≠-1
@籍重6059：求方程(x+1)y''+y'=ln(x+1)的通解时,若令y'=p,则: - 作业帮
奚怕14796509783…… [选项] A. y''=p' B. y''=p(dp)/(dy) C. y''=p(dp)/(dx) D. y''=p'(dp)/(dy)中那个选项正确,为什么?
@籍重6059：函数z=1/ln(x+y+1)的定义域 -
奚怕14796509783…… ln(x+y+1)≠0 【它充当分式的分母,当然不能为0】也就是ln(x+y+1)≠0 =ln1 x+y+1≠1且x+y+1>0【对数的真数必须大于0】联合得到:x+y∈(-1,0)∪(0,+∞)
@籍重6059：Y=(X+1)LN(X+1)求导?要过程. -
奚怕14796509783…… 乘法法则 y=(x+1) ln(x+1) y'=(x+1)' ln(x+1)+(x+1)[ln(x+1)]' =ln(x+1)+1
@籍重6059：求(x+1)y''+y'=ln(x+1)的通解? -
奚怕14796509783…… 解:令x+1=e^t,则(x+1)y'=dy/dt,(x+1)²y''=d²y/dt²-dy/dt 代入原方程得d²y/dt²-dy/dt+dy/dt=te^t ==>d²y/dt²=te^t ==>dy/dt=∫te^tdt=te^t-e^t+C1 (C1是积分常数) 于是,y=∫[(te^t-e^t+C1]dt =te^t-2e^t+C1t+C2 (C2是积分常数) =(x+1)ln(x+1)-2(x+1)+C1ln(x+1)+C2 故原方程的通解是y=(x+1)ln(x+1)-2(x+1)+C1ln(x+1)+C2 (C1,C2是积分常数)
@籍重6059：xy=ln(x+y)求导数步鄹 -
奚怕14796509783…… 两边对x求导,得 y+xy'=1/(x+y) ·(1+y') (x-1/(x+y))y'=1/(x+y)-y 所以 y'=【1/(x+y)-y】/(x-1/(x+y))
