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Consider the IVP \begin{equation*}\tag{1} x''+p(t)x'+q(t)x=r(t) \end{equation*} where $p,q,r$ are continuous functions on $\mathbb{R}$. Suppose $t,t^2+1,t^2+2t$ are solutions of the IVP on the interval $I$ containing $0$.

  1. What is the fundamental set of solutions of the homogeneous equation and
  2. find the solution to $(1)$ on $I$ which satisfies the initial conditions $x(0)$ and $x'(0)=2$.

I need some explanation if you got the solution. I think i may miss some important concept about DE.

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Hint: if $x_1(t)$ and $x_2(t)$ are solutions to your equation then $x_1(t)-x_2(t)$ is a solution to the homogeneous one (i.e., which has $r(t)\equiv 0$). You should know that a second order linear differential equation has two dimensional solution space, i.e., you need to find two linearly independent solutions to the homogeneous equation, this will be your fundamental set of solutions. –  Artem Apr 29 '12 at 11:48
    
So $p(t)+tq(t)=r(t)$, $2+2tp(t)+(t^2+1)q(t)=r(t)$ and $2+(2t+2)p(t)+(t^2+2t)q(t)=r(t)$ for all $t$. Maybe you can get some information about $p$, $q$ and $r$. –  Siminore Apr 29 '12 at 11:51
    
@Artem, I dont quite get how you conclude x1(t) -x2(t) would give the solution to the homogeneous equation –  Mathematics Apr 30 '12 at 1:31
    
@Mathematics, If $Lx_1(t)=r(t), Lx_2(t)=r(t)$, where $L$ is a linear differential operator, then, using the linearity, $0=Lx_1(t)-Lx_2(t)=L(x_1(t)-x_2(t))$, hence $x_1-x_2$ solves the homogeneous equation. –  Artem Apr 30 '12 at 4:23
    
@Mathematics $(1)$ isn't an IVP yet you say it is. What do you really want to say? –  Git Gud Nov 23 '13 at 17:28

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