解(1)分离变量 (1+x)/xdx=-(1-y)/ydy x y 两边积分 lnx+x=-lny+y+c 所以,原方程的通解为 x-y+ln(xy)=c (c为任意常数). 注意:在求解微分方程过程中,为方便起见,且不失一般 性,遇到积分 ∫1/udx ,都假设u0,即∫1/udx=lnu+c (2)原方程可化为 x(y^2-1)dx=-y(1-x^2...
1. (1) 可取 μ=e^(3x) , 得通积分为 3x^2y+y^3=Ce^(-3x) ; (2)可取 μ=1/ye^(2y) ,得通积分为 xe^(2y)-ln|y|=C 特解 y = 0; (3)可取 μ=xy , 得通积分为 x^3y+3x^2+y^3=C ; (4)可取 μ=1/(x^2+y^2) ,得通积分为 y+arctany/x=C, 特解 ...
Question: ∫C(ydx+xdy) C x=t3,y=t2,2≤t≤4. Vector Line Integrals: LetCbe an oriented curve parameterized byr(t)fora≤t≤b, and letF→be a vector field whose domain includesC. Then the line integral ofF→on...
Suppose F(x,y) = (x + 5)i + (3y + 3)j. Use the fundamental theorem of line integrals to calculate the following: The line integral of Compute the following line integral: integral over C of (x + y)dx + (xy)dy when C is the broken line runn...
设函数f(x,y)具有一阶连续偏导数,且(x,y)=yedx+x(1+y)edy,f(0,0)=0,则f(x,y)=___. 查看答案进入试题列表
3.求下列微分方程的通解(1) (e^(x+1)-e^x)dx+(e^(x+y)+e^y)dy=0(2) ydx+(x^2-2x)dy=0 .(3) (x-ycosy/x)dx+xcosy/xdy=0 .(4) xdy-ydx=y^2e'dy .(5) y'+1/xy=e^(-x^2) .(6) y'=+cosx.colx(7) y''-7y'+12y=0 .(8) 4y''-12y'+9y=0 .(...
(x^2-2x)dy=03) (x-ycosy/x)dx+xcosy/xdy=0.(4) xdy-ydx=y^2e^ydy(5) y'+1/xy=e^(-x^2)(6) y'=y/(cotx)+cosx.cotx(7) y''-7y'+12y=0.(8) 4y''-12y'+9y=0(9) y''+2y'+3y=0.(10) y''-2y'+3y=x^2-x+2(11) y''+2y'-3y=e^(2x)(12) y''...
1LerR8mbFquqsTWjt8XyE6Q9NQp16AgQAW 0.00100847BTC 1Af5e47P36tmW1keZWn8XEKVejT9FEw26P 0.00100847BTC 1MqA4PZGYq7FikmX9Ex1qarE5wSBjXzBd6 0.00100847BTC 1Ak8g2KeCSNyFQBHVAL5zGjGXYX1Xi6rDJ 0.00100847BTC查看全部 (剩余 490 条) Bybit. DepositAndWithdraw_2(1Grw...kmqC) 0.50353072BTC 区块:88700...
nPTEmb3Blbl9jYXRlZ29yeV93aGVuX25vdF9hZGRlZD0yJnpsaW5rPWh0dHBzJTNBJTJGJTJGYXJ0aWNsZS56bGluay50b3V0aWFvLmNvbSUyRmg2NWMmemxpbmtfY2xpY2tfdGltZT0xNjE2OTEwMzI2ODAxJnpsaW5rX2RhdGE9JTdCJTIycG9zaXRpb24lMjIlM0ElMjJyZWNvbW1lbmRlZCUyMiUyQyUyMmdpZCUyMiUzQSUyMjE2OTU0NDE4MzA2MTE5ODMlMjIlM...
"image":"https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjiRsXdZy_2eIKbE6hxzfPxkEFYdmRVCYWyRXqGaIpLxG_Y-av8Dxz8sImw9b6LS61t4Q38Ykbdy2dR1MUG5U0dAu2ayPMPASm3LAMrNlywqy_ZgeUaDCCywKcR8ihVBwCEYFQ-Igxv8dEVhkcu480fG2NQSqWai0Dqbp4PtJVA3Os_ewmY5sNKblBXbQ/s1600/00-korean...