Question

An Interesting Equation of Order n: xy" - (x + n)y' + ny = 0: u(x) = e^x. (a) Prove that the given solution is indeed a solution. (b) Obtain a second, linearly independent, solution.

Answer 1

The given differential equation is

x y" -(x+n)y'+ny=0------------(1)

Let, y'=t

y"=t'

`(dy)/(dx)=t\\\\y=t x`

Substituting the value of , y', y'' and y in equation (1)

→x t' -(x+n)t+n t x=0

→ x t' = x t+ n t- nt x

→ x t'=t(x+n-nx)

`\rightarrow (t')/(t)=(x+n-nx)/(x)\\\\\rightarrow (dt)/(t)=(1-n+(n)/(x)) dx\\\\ \text{Integrating both sides}}\\\\ \int{(dt)/(t)}=\int {(1-n+(n)/(x)) dx}\\\\ \log t=x - nx+n \log x+\log K\\\\ \log t -\log x^n-\log K=x(1-n)\\\\\log (t)/(Kx^n)=x(1-n)\\\\t=Kx^n * e^(x(1-n))\\\\y'=Kx^n * e^(x(1-n))`

which is a solution of Differential equation.

(b)

`(dy)/(dx)=Kx^n* e^(x(1-n))\\\\ dy=Kx^n* e^(x(1-n)) dx`

Integrating both sides

`y=(Kx^n* e^(x(1-n)))/(1-n)-(Kn)/(1-n)\int{x^(n-1)e^(x(1-n)) dx`

required linear independent solution of Differential equation.