91影视

It has direct application to Fourier's approach in the study of partial differential equations, the Cauchy-Euler equation is significant in the theory of linear differential equations.

It is given that for the differential equation

ay"+by'+cy=0

Where a, b, c are constants, the characteristic equation is,

ar²+br+c=0

For which there is a repeated root at r₀ and one solution of the equation is

y₁=e^r₀t

Using operator approach, if r₀ is a repeated root, then

L[w] (e^t) = a (r-r₀)² e^rt

You observe that, the righthand side of the above equation has a factor (r-r₀)², so taking the partial derivative, w.r.t. r and setting r=r₀ you will get 0,

饾浛 /饾浛r [L [w] (t)] |_r=r0= [a(r-r₀)² (re^rt)+2a (r-r₀) e^rt]|_r=r0=0

You also note that w(r,e^t)=e^rt has continuous partial derivatives of all orders with respect to both and hence the mixed partial derivates are equal.

饾浛²飞/饾浛谤饾浛迟² = 饾浛²飞/饾浛谤²饾浛t ,

饾浛²飞/饾浛谤饾浛迟 = 饾浛²飞/饾浛谤饾浛迟

Consequently, for a differential operator L, you have,

饾浛/饾浛r L[w] = 饾浛/饾浛r {a 饾浛²飞/饾浛迟² +b 饾浛飞/饾浛迟+cw}

=补饾浛³飞/饾浛谤饾浛迟² +b 饾浛²飞/饾浛谤饾浛迟 +c 饾浛飞/饾浛谤

=补饾浛³飞/饾浛迟²饾浛r +b 饾浛²飞/饾浛迟饾浛r +c 饾浛飞/饾浛谤

=L [饾浛飞/饾浛谤]

From above, you can say that the operators 饾浛/饾浛r and L are commutative. Thus,

L [饾浛飞/饾浛谤]|_r=r0=0

Therefore, in the case of repeated roots at r₀, the second linearly independent solution is,

y₂(x) = 饾浛飞/饾浛谤 (r₀ , e^t)

=饾浛/饾浛r (e^rt) |_r=r0

=te^r₀t

Hence, you find the second solution for the differential equation is derived to be y₂=te^r₀t where the first solution is given to be y₁=e^r₀t .