设x=e^(-t),变换方程x^2*d^2y/dx^2+x*dy/dx+y=0
设x=e^(-t),变换方程(x^2)*d^2y/dx^2+x*dy/dx+y=0
答案是d^x/dt^2+y=0
求解啊~~
dy/dx = (dy/dt) / (dx/dt) = (-1/x) * dy/dt
d²y/dx² = (1/x²) * dy/dt + (-1/x) * d²y/dt² / (dx/dt) = (1/x²) * dy/dt + (1/x²) * d²y/dt²
= (1/x²) * [ dy/dt + d²y/dt² ]
x * dy/dx = - dy/dt, x² * d²y/dx² = dy/dt + d²y/dt²
代入原方程，得：d²y/dt² + y= 0
设x=e^(-t),变换方程x^2*d^2y\/dx^2+x*dy\/dx+y=0
x * dy\/dx = - dy\/dt, x² * d²y\/dx² = dy\/dt + d²y\/dt²代入原方程，得：d²y\/dt² + y= 0
d²y\/dx²+2dy\/dx+y=e∧x
x * dy\/dx = - dy\/dt,x² * d²y\/dx² = dy\/dt + d²y\/dt²代入原方程,得：d²y\/dt² + y= 0
设x=e^(-t) 试变换方程x^2 d^2y\/dx^2 +xdy\/dx+y=0
所以x^2 d^2y\/dx^2= d^2y\/dt^2 +dy\/dt，而xdy\/dx= -dy\/dt，于是原方程可以变换为：d^2y\/dt^2 +y=0
设x=e^(-t) 试变换方程x^2 d^2y\/dx^2 +xdy\/dx+y=0
-e^t dy\/dt](-e^t)=e^(2t)d^2y\/dt^2 +e^(2t)dy\/dt 所以x^2 d^2y\/dx^2= d^2y\/dt^2 +dy\/dt，而xdy\/dx= -dy\/dt，于是原方程可以变换为:d^2y\/dt^2 +y=0
x=e^(-t);y=te^t 则d^2y\/dx^2=
x=e^(-t);y=te^t所以dx\/dt=-e^(-t)dy\/dt=e^t+te^tdy\/dx=(1+t)e^t\/(-e^(-t))=-(1+t)e^2td(dy\/dx)\/dt=-e^(2t)-2(1+t)e^2t=-(2t+3)e^2t所以d^2y\/dx^2=【-(2t+3)e^2t】\/【-e^(-t)】=(2t+3)e^3t ...
x=e^(-t);y=te^t 则d^2y\/dx^2=
x=e^(-t);y=te^t所以dx\/dt=-e^(-t)dy\/dt=e^t+te^tdy\/dx=(1+t)e^t\/(-e^(-t))=-(1+t)e^2td(dy\/dx)\/dt=-e^(2t)-2(1+t)e^2t=-(2t+3)e^2t所以d^2y\/dx^2=【-(2t+3)e^2t】\/【-e^(-t)】=(2t+3)e^3t ...
微分方程求解答谢谢
令x=e^t，则t=lnx，dt\/dx=1\/x dy\/dx=dy\/dt*dt\/dx=(1\/x)*dy\/dt d^2y\/dx^2=d(dy\/dx)\/dx=d[(1\/x)*dy\/dt]\/dt*dt\/dx=(1\/x^2)*(d^2y\/dt^2-dy\/dt)代入原方程 d^2y\/dt^2-3dy\/dt+2y=1 特征方程r^2-3r+2=0，r1=1，r2=2 齐次方程的通解为y$=C1*e^t+C2*e...
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y=f(lnx) 求(d^2y)\/(dx^2)
dy\/dx = 2xf'(x^2)d^2y\/dx^2 = d(2xf'(x^2))\/dx = 2f'(x^2)+ 4x^2f''(x^2)这些都是套用复合函数导数公式而已，lz应该能自己搞出来 求采纳为满意回答。
...^t ,dy\/dx=dy\/xdt .分析变换具体步骤 d^2y\/dx^2=(d^2y\/dt^2-dy\/...
x=e^t ,dy\/dx=dy\/xdt dt\/dx=1\/x
