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.

Solution:

The given differential equation is

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

Let, y'=t

y"=t'

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)

which is a solution of Differential equation.

(b)

Integrating both sides

required linear independent solution of Differential equation.

f(x) = y

f(-3) = -5

step-by-step explanation:

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.