What's the dimension of the solution space of $x^2y'' − 4xy' + 6y = 0$ on $(−1, 1)$?
The answer is $3$.
How is this possible? Isn't the basis $\{x^2 , x^3\}$?
What's the dimension of the solution space of $x^2y'' − 4xy' + 6y = 0$ on $(−1, 1)$?
The answer is $3$.
How is this possible? Isn't the basis $\{x^2 , x^3\}$?
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What about this solution? $$ y(x) = \begin{cases} x^3, & x \ge 0 \\ -x^3, & x < 0 \end{cases} $$