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

Solution:

The given differential equation is

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

Let, y'=t

y"=t'

$\frac{dy}{dx}=t\\\\y=t x$

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)

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

which is a solution of Differential equation.

(b)

$\frac{dy}{dx}=Kx^n\times e^{x(1-n)}\\\\ dy=Kx^n\times e^{x(1-n)} dx$

Integrating both sides

$y=\frac{Kx^n\times e^{x(1-n)}}{1-n}-\frac{Kn}{1-n}\int{x^{n-1}e^{x(1-n)} dx$

required linear independent solution of Differential equation.