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It's seems like a really simple question, but I can't understand how to solve it.

I am requested to decide which function $y=t^2$, or $y=t^2+1$ can be used for a particular solution of an order two equation: $y''+a(t)y'+b(t)y=0$ with the continues coefficients in all $\mathbb R$. I need to give the equation as well.

What should I do? usually I am asked to decide between two pairs, so I put each pair in it's Wronskian matrix and check if it's determinant can't be $o$ in all the given range. here $y=t^2+1$ is not $0$, but shouldn't I check its derive? and then usually I put both the pair and $y$ in a Wronskian matrix and find the equation, what should I do here?


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How about taking the derivatives of $y$, plugging them into the equation, and see what it tells you? – Ross Millikan Jul 4 '12 at 16:18
I did it. it tells me nothing. – Jozef Jul 4 '12 at 16:19
@Jozef If your coefficients $a$ and $b$ are not explicitly given, the question becomes a bit funny :-) Please, tell us: do you have a particular ODE? – Siminore Jul 4 '12 at 16:30
it $can be used$ as a particular solution for a general equation of the above form. it's not that funny. – Jozef Jul 4 '12 at 16:45
I was confused about the language: a function either is or isn't a solution for a given ODE. I did not understand that the problem was to find an ODE which is solved by a given function. It is an inverse problem, so to speak. – Siminore Jul 4 '12 at 17:14
up vote 3 down vote accepted

Plug $y=t^2$ and $y=t^2+1$ into the equation and set $t=0$. What do you get?

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  • $y=t^2$, $y'=2t$ and $y''=2$:

    $2+a(t) \cdot 2t + b(t) \cdot t^2=0\tag{1}$

  • $y=t^2+1$, $y'=2t$ and $y''=2$:

    $2+a(t) \cdot 2t + b(t) \cdot (t^2+1)=0\tag{2}$

Solve the system of equations ($(1)$ and $(2)$) for $a(t)$ and $b(t)$, you get $b(t)=0$ and $a(t)=-\frac 1t$.

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Thank you for the answer. I get that the $y=t^2$ with $t=0$ yields that $2=0$, so it has to be the other equation. If $y=t^2$ cannot be used, so why can you use it in order to solve the above system? – Jozef Jul 4 '12 at 16:44
If using $t^2+1$, there are infinitely many $a(t)$, $b(t)$ for which $t^2+1$ is a particular solution. – André Nicolas Jul 4 '12 at 17:06

Consider $y(t)=t^2$. Then $y'(t)=2t$ and $y''(t)=2$. The function $y$ is a solutions iff $$ 2+2ta(t)+t^2 b(t)=0 \qquad \forall t. $$ Exercise: do the same for $y(t)=t^2+1$ :-)

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