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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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    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**.2012-11-09

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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:

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