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Given that $t,1+t,t^2,-t$ are the solutions to $y'''+a(t)y''+b(t)y"+c(t)y=d(t)$,what is the solution of homogeneous equation to this differential equation? What i have done is tried the properties of linear differential equation that $L(t)=L(1+t)=L(t^2)=L(-t)=d(t)$ so the homogeneous solution should be independent and i claim that $1,t,t^2$ should be the solution. However, i am not sure hot can i actually conclude that these are the solutions? It seems that it can be quite a number of sets of solution by the linearity.

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2 Answers 2

up vote 3 down vote accepted

Substitute the four given solutions into the original equation, which I assume is supposed to be $y'''+a(t)y''+b(t)y'+c(t)y=d(t)$:

$$\begin{array}{rl} y=t:&b(t)+tc(t)=d(t)\\ y=-t:&-b(t)-tc(t)=d(t)\\ y=t+1:&b(t)+(1+t)c(t)=d(t)\\ y=t^2:&2a(t)+2tb(t)+t^2c(t)=d(t) \end{array}$$

Now add the first two equations in the righthand column, and you see that the equation already is homogeneous, and any linear combination of these four solutions is a solution of the homogeneous equation. In particular, $y=\alpha+\beta t+\gamma t^2$ is a solution for all $\alpha,\beta,\gamma\in\Bbb R$.

Added: You can also use the approach mentioned by anon in the comments to justify your answer: $t-(-t)=2t$ is a solution to the homogeneous equation, so $t$ is, and then $t^2-t$ is a solution to to the homogeneous equation, so $(t^2-t)+t=t^2$ is.

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So, you get that $d(t)=0$. Since $1$ is a solution, we get $c(t)=0$. Since $t$ is a solution, $b(t)=0$. Since $t^2$ is a solution, $a(t)=0$. –  robjohn May 4 '12 at 9:13
    
@robjohn: Good point. Do you want me to add it to the answer, or shall we just leave it as a comment? –  Brian M. Scott May 4 '12 at 9:19
    
If you want to incorporate it, I see no reason not to. Of course, the equation then becomes: $y^{\,\prime\prime\prime}=0$. –  robjohn May 4 '12 at 9:33

There are actually an infinite number of solutions to the homogeneous version of the differential equation. You'll notice that $y=0$ is a solution, and for any two solutions $u$ and $v$,

$$L(\alpha u+\beta v)=\alpha Lu+\beta Lv=0,$$

so linear combinations of solutions are also solutions. Therefore the solutions form a vector space, and it is a fact that the dimension of the space is equal to the order of the differential equation, which is three. To characterize the space it suffices to find three linearly independent solutions so that they can be used as a basis for the space (every other solution will be a combination of the basis solutions). Of course $\{1,t,t^2\}$ is LI and is a very nice choice of polynomials for a basis.

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Clear explaination,thx! –  Mathematics May 4 '12 at 9:13

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