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Thread: Clarification of Problem

  1. #1
    Junior Member
    Oct 2012

    Clarification of Problem

    I have just completed a section on Second ODE and Second Order Linear Homogenous equations. Topics were the superposition principle, the wronskian, and linear dependence and independence.

    But Part A of this exercise question is not clear to me.

    Given the diff equ $\displaystyle (x^2+2x-1)y"-2(x+1)y'+2y=0$

    a) Show that the equation has a linear polynomial and a quadratic polynomial as solutions. ??
    b) Find two linearly independent solutions of the equation and find the general solution. (For this do i assume the answer is in the form $\displaystyle y=x^r$)

    Any clarification to help get me started is appreciated. Thanks.
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  2. #2
    MHF Contributor
    Nov 2013

    Re: Clarification of Problem

    looking at various solutions I'm pretty sure you mean $\displaystyle (x^2+2x+1)y''$, not what you have written.

    a linear polynomial is $\displaystyle pl(x)=ax+b$

    a quadratic polynomial is $\displaystyle pq(x)=a x^2+b x + c$

    first step should just be plugging these in and confirming that they are solutions.

    For example for the linear polynomial

    $\displaystyle y(x)=ax+b$

    $\displaystyle y'(x)=a$

    $\displaystyle y''(x)=0$

    so plugging these in we get

    $\displaystyle (x^2+2x+1) \cdot 0 - 2(x+1)\cdot a + 2(ax+b)=0$

    $\displaystyle -2ax-2a+2ax+2b=0$

    $\displaystyle a=b$

    so any linear polynomial of the form ax+a is a solution. This can be thought of as a(x+1), i.e. (x+1) serves as a basis vector of the solution space.

    Now repeat this with the quadratic. and remember superposition, and then you'll start to understand what they are after in part 2 of the problem.
    Last edited by romsek; Feb 11th 2014 at 09:35 PM.
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