y''-x[SUP]2[/SUP]y=0, (y''=d[SUP]2[/SUP]y/dx[SUP]2[/SUP])

when try to find solution, usually assuming solution form is y=e[SUP]lambda X[/SUP], and lambda =x, -x, then general solution is y=c[SUB]1[/SUB]e[SUP]x^2[/SUP] + c[SUB]2[/SUB]e[SUP]-x^2[/SUP]. however, I was told this is wrong approach to solve.

Then how can I get correct general solution?

when try to find solution, usually assuming solution form is y=e[SUP]lambda X[/SUP], and lambda =x, -x, then general solution is y=c[SUB]1[/SUB]e[SUP]x^2[/SUP] + c[SUB]2[/SUB]e[SUP]-x^2[/SUP]. however, I was told this is wrong approach to solve.

Then how can I get correct general solution?

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