作业帮设x=e^(-t) 试变换方程x^2 d^2y/dx^2 +xdy/dx+y=0
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设x=e^(-t) 试变换方程x^2 d^2y/dx^2 +xdy/dx+y=0
x=e^(-t),即dx/dt= -e^(-t)
那么dy/dx=(dy/dt) / (dx/dt)= -e^t *dy/dt,
而
d^2y/dx^2
= [d(dy/dx) /dt] * dt/dx
= [-e^t *d^2y/dt^2 -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
那么dy/dx=(dy/dt) / (dx/dt)= -e^t *dy/dt,
而
d^2y/dx^2
= [d(dy/dx) /dt] * dt/dx
= [-e^t *d^2y/dt^2 -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
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