于是，代入原方程得d²y/dt²-2dy/dt+(dy/dt)²=0........(1)
再令dy/dt=p，则d²y/dt²=dp/dt
于是，代入方程(1)得dp/dt-2p+p²=0
==>dp/(p(2-p))=dt
==>ln│p/(2-p)│=ln│2t│+ln│C1│ (C1是积分常数)
==>p/(2-p)=C1e^(2t)
==>dy/dt=p=2-2/(C1e^(2t)+1)
==>y=2t+ln│C1+e^(-2t)│+C2 (C2是积分常数)
==>y=2ln│x│+ln│C1+1/x²)│+C2
==>y=ln│C1x²+1│+C2
经验证y=C (C是积分常数)也是原方程的解
故 原方程的所有解是y=ln│C1x²+1│+C2，或y=C (C,C1,C2是积分常数)
求xy''-y'+xy'^2=0的通解
解：令x=e^t，则xy'=dy\/dt，x²y''=d²y\/dt²-dt\/dt 于是，代入原方程得d²y\/dt²-2dy\/dt+(dy\/dt)²=0...(1)再令dy\/dt=p，则d²y\/dt²=dp\/dt 于是，代入方程(1)得dp\/dt-2p+p²=0 ==>dp\/(p(2-p))=dt ==>ln│...
y'^2+xy'-y=0的通解
有什么问题请追问
微分方程xy'-y-根号下(x^2+y^2)=0的通解。
x=r*cosθ y=r*sinθ 极坐标系下dx=cosθ dr-sinθ rdθ dy=sinθ dr+cosθ rdθ 方程化为rcosθ *(sinθ dr+cosθ rdθ)\/(cosθ dr-sinθ rdθ ) -r*sinθ-r=0 化简为dr\/r = dθ*(1+sinθ)\/cosθ 右边=dθ* cosθ*(1+sinθ)\/(cosθ)^2=d(sinθ)\/(1-...
求微分微分方程通解 xy'—y—(x^2十y^2)^0.5=0
解：∵xy"=y'-xy'^2 ==>xdy'\/dx=y'-xy'^2 ==>y'dx-xdy'-xy'^2dx=0 ==>(y'dx-xdy')\/y^2-xdx=0 ==>∫(y'dx-xdy')\/y^2-∫xdx=0 (积分)==>x\/y'-x^2\/2=C1\/2 (C1是常数)==>y'=2x\/(x^2+C1)∴y=∫2xdx\/(x^2+C1)=∫d(x^2+C1)\/(x^2+C1)=...
一阶线性微分方程xy'-y-√x²+y²=0求通解
这是一阶齐次微分方程 (x^2+y^2)dx-xydy=0 dy\/dx=(x²+y²)\/(xy)dy\/dx=((x\/y)²+1)\/(x\/y)令u=y\/x 则dy=du*x+dx*u dy\/dx=(du\/dx)*x+u 代入得 (du\/dx)*x+u=(u²+1)\/u=u+1\/u du\/dx=1\/(xu)u*du=dx\/x 两边积分得 (1\/2)u²...
微分方程xy''-y'+2=0的通解?
简单计算一下即可，答案如图所示
求xy'-y-√(x^2+y^2)=0的通解,请写过程,方法越简单越好
dy\/dx=y\/x+√[1+(y\/x)^2]令u=y\/x,则y=xudy\/dx=u+xdu\/dx∴u+xdu\/dx=u+√(1+u^2)∴xdu\/dx=√(1+u^2)∴du\/√(1+u^2)=dx\/x∴ln[u+√(1+u^2)]=lnx+C1∴u+√(1+u^2)=Cx【其中C=e^C1】√(1+u^2)=Cx-u∴1+u^2=(Cx)^2+u^2-2Cxu∴1=(Cx)^2-2Cy∴通解为(...
求y'+xy^2=0的通解,满足初始条件y(0)=2的特解
∵y'+xy²=0 ==>dy\/dx=-xy²==>dy\/y²=-xdx ==>-1\/y=-x²\/2-C\/2 (C是任意常数)==>(x²+C)y=2 ∴原方程的通解是(x²+C)y=2 ∵y(0)=2 ∴C=2 故 原方程在初始条件y(0)=2下的特解是(x²+2)y=2.
求xy'-y-e^x*y^2=0的通解
解：显然，y=0是原方程的解 当y≠0时，∵令y=1\/z，则y'=-z'\/z^2 代入原方程，化简得 xz'+z+e^x=0 ==>xdz+zdx+e^xdx=0 ==>d(xz)+d(e^x)=0 ==>xz+e^x=C (C是常数)==>x\/y+e^x=C ∴x\/y+e^x=C也是原方程的解 故原方程的通解是y=0和x\/y+e^x=C。
xy'-y-y*y=0的通解
解：∵xy'-y-y²=0 ==>xdy\/dx=y(y+1)==>dy\/[y(y+1)]=dx\/x ==>[1\/y-1\/(y+1)]dy=dx\/x ==>ln│y│-ln│y+1│=ln│x│+ln│C│ (C≠0是积分常数)==>y\/(y+1)=Cx ∴原微分方程的通解是y\/(y+1)=Cx (C≠0是积分常数)。
