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Verify that $y_1(t) = t^2$ and $y_2(t) = t^{-1}$ are two solutions of the differential equation $t^2y^{''}-2y = 0$ for $t>0$. Then show that $c_1t^2 +c_2t^{-1}$ is also a solution of this equation for any $c_1$ and $c_2$.

For the first part, I know it is a solution since if I plug in $y_1(t) = t^2$ and $y_2(t) = t^{-1}$ into the differential equation it will give me $0$ which is the solution to the differential equation, but how can I do the second part. Showing that $c_1t^2 +c_2t^{-1}$ is a solution? If I plug those in the differential equation I get something that isn't equal to zero.

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Then you made a little slip, easy to do, there are a lot of minus signs around! Note that the second derivative of $t^{-1}$ is $\frac{2}{t^3}$. – André Nicolas Nov 9 '12 at 6:57
@AndréNicolas I feel dumb now. I was going at this question for 20 mins. I should note it is pretty late and I should go to sleep. Thanks a lot! I am going to give the answer below just for future users for reference. – Q.matin Nov 9 '12 at 7:02
Good idea, it is best if questions do not remain unanswered. Note that the part with the constants is an almost automatic no computation consequence of the fact the two given solutions are solutions, because the DE is linear. – André Nicolas Nov 9 '12 at 7:43
up vote 1 down vote accepted

First part of question $y_1(t) = t^2$ and $y_2(t) = t^{-1}$ are solutions since if we plug it into the differential equations we get:

$$(t^2)^{''} - \frac{2}{t^2}(t^2) = 2 -2 = 0$$

$$(t^{-1})^{''} - \frac{2}{t^2}(t^{-1}) = \frac{2}{t^3} - \frac{2}{t^3}=0$$

For the second part:

$$(c_1t^2 +c_2t^{-1})^{''} - \frac{2}{t^2}(c_1t^2 +c_2t^{-1}) = 2c_1 + \frac{2}{t^3} - 2c_1 -\frac{2}{t^3}=0$$

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