Finding the homogenous solution of Var.Coeff. 2nd Order ODE

[dumbdumNotSmart]

It's been too long guys. I've given this ODE lots of thought and still no cigar.

Homework Statement

We are given the following ODE:
$$ (x-a)y''-xy'+a^2y = a(x-1)^2e^x $$
and knowing that y=e^x is a solution to the homogenous equation, find the possible values of a.

Next part: Using the obtained values, calculate the solution (y(x)) so that it's limit as x→ -∞ is contained within a finite set.
(Also, let the solution be continuous in said set)

Homework Equations

The Attempt at a Solution

First I replaced the known base of the homogenous solution in the equation. I got that a can be either 0 or 1.

From here I investigate for the case if a=1
Knowing a base of the homogenous solution, I set out to find the missing one knowing a ODE of order 2 has 2 bases for the Homogenous solution. The way I tried to find it was writing it in the form of y₂=α(x)e^x deriving it 2 times, then replacing each expression in it's respective place in the homogenous ODE. Using the principle of superposition I rearranged the resulting expression as follows:

$$ e^x \left( x(\alpha ''+\alpha ' )-( \alpha '' +2\alpha ') \right)=0 $$

Where α is an unknown function. From here we get two seemingly contradictory equations, one points to α being e^-x and the other to e^-2x, each giving me a different base of the homogenous equation. This cannot be, I must be doing something wrong. But what?

 

[ehild]

Homework Statement

We are given the following ODE:
$$ (x-a)y''-xy'+a^2y = a(x-1)^2e^x $$
and knowing that y=e^x is a solution to the homogenous equation, find the possible values of a.

Next part: Using the obtained values, calculate the solution (y(x)) so that it's limit as x→ -∞ is contained within a finite set.
(Also, let the solution be continuous in said set)

Homework Equations

The Attempt at a Solution

First I replaced the known base of the homogenous solution in the equation. I got that a can be either 0 or 1.

From here I investigate for the case if a=1
Knowing a base of the homogenous solution, I set out to find the missing one knowing a ODE of order 2 has 2 bases for the Homogenous solution. The way I tried to find it was writing it in the form of y₂=α(x)e^x deriving it 2 times, then replacing each expression in it's respective place in the homogenous ODE. Using the principle of superposition I rearranged the resulting expression as follows:

$$ e^x \left( x(\alpha ''+\alpha ' )-( \alpha '' +2\alpha ') \right)=0 $$

Where α is an unknown function.

e^x is never zero, so you have the de for α(x) : α''(x-1)+α'(x-2)=0 Solve for α' first, then integrate to get α(x).

From here we get two seemingly contradictory equations, one points to α being e^-x and the other to e^-2x, each giving me a different base of the homogenous equation. This cannot be, I must be doing something wrong. But what?

Your solutions are not correct. There is no contradiction.

 

[dumbdumNotSmart]

I redid my calculations. Principle of superposition is not valid for findibg the homogenous since alpha is a function. Still confused on one point. If the homogenous solution I get is a sum of x and 2, its like you have two bases on top of what they gave you. Did I do something wrong?

 

[ehild]

I redid my calculations. Principle of superposition is not valid for findibg the homogenous since alpha is a function. Still confused on one point. If the homogenous solution I get is a sum of x and 2, its like you have two bases on top of what they gave you. Did I do something wrong?

2 is not solution of the homogeneous equation. Have you plugged in y=2?