Second order DE from my midtearm

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Homework Help Overview

The discussion revolves around a second-order ordinary differential equation (ODE) of the form y'' + p(x)y' + q(x)y = g(x). The original poster notes that specific functions, y = x², y = x, and y ≡ 1, are solutions to this equation and seeks to find the general solution.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to derive the general solution based on the given particular solutions, questioning the validity of their approach to linear independence and the principle of superposition. They also explore the idea of singular solutions and express uncertainty about selecting a particular solution for the non-homogeneous equation.

Discussion Status

Participants have engaged in clarifying the nature of the solutions and the general solution's structure. Some guidance has been provided regarding the relationship between particular solutions and the general solution, with an emphasis on the need to consider the homogeneous aspect of the equation. Multiple interpretations of the general solution are being explored.

Contextual Notes

The original poster is working within the constraints of a midterm exam question, which may impose specific expectations on the formulation of the general solution. There is an ongoing discussion about the appropriateness of using particular solutions in the context of a non-homogeneous equation.

Danny B
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Homework Statement



Consider the ODE:

y''+p(x)y'+q(x)y=g(x)

It is given that the functions y=x[itex]^{2}[/itex], y=x and y[itex]\equiv[/itex]1
are solutions of the equation.
Find the general solution of the equation.

Homework Equations



The Attempt at a Solution



Well, given the three solutions, and taking into account that they are mutually linearly independent, I would say that any couple of these functions will give me a set that forms the general solution.

If i take y=x and y[itex]\equiv[/itex]1, I get the general solution:

y(x)=c1+c2*x

Now, I know that x[itex]^{2}[/itex] is also a solution, but it can't be derived from the general solution, so obviously I'm doing something wrong.

I also thought of the possibility that y=1 is a singular solution and the general solution is:

y(x)=c1*x+c2*x^2

does this makes sense?

This is a question from my midterm and i still can't figure it out. So... help?
 
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Welcome to PF!

Hi Danny B! Welcome to PF! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)

Each of those three are particular solutions …

Hint: what relation is there between any two particular solutions (ie any two solutions)? :wink:
 
Thank you.

I think I got it now.

My mistake was to use the principle of superposition with a non-homogeneous equation.

So the general solution should be:

(Considering that any difference of the particular solutions is a homogeneous solution)

y[itex]_{(x)}[/itex]=c[itex]_{1}[/itex](x-1)+c[itex]_{2}[/itex](x[itex]^{2}[/itex]-1)+Y[itex]_{p}[/itex]

But how do I choose Y[itex]_{p}[/itex]?

Can I just use any of the three particular solutions?
 
Hi Danny B! :smile:

Let's write it out in full, so you can see how it works …

the general solution can be written in three ways

A(x2 - 1) + B(x - 1) + 1

A(x2 - x) + x + B(1 - x)

x2 + A(x - x2) + B(1 - x2)​

the homogeneous general solution can be written in three ways

A(x2 - 1) + B(x - 1)

A(x2 - x) + B(1 - x)

A(x - x2) + B(1 - x2)​

now you need to spend ten minutes convincing yourself that they're the same! :rolleyes:

he he :biggrin:
 
Thanks a lot!

You've been a great help!
 

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