(adsbygoogle = window.adsbygoogle || []).push({}); Consider the equation 2t^2y'' + 3ty' - y = 0

(a) Show that y1(t) = sqrt(t) and y2(t) = 1/t are soltuions of the equation on the interval 0<t<infinity

(b) Compute W[y1,y2](t). What happens as t approaches zero?

(c) Show that y1(t) and y2(t) form a fundamental set of solutions of the equation on the interval 0<t<infinity.

For (a), i just did lhs rhs and proved they equaled, i did not do anything with the interval, am i suppose to?

For (b) I computed it to be W=-3/(2t^(3/2)) which is the correct answer. and i got as t->0, W-> - infinity, is this correct? can someone confirm.

For (c) i'm not sure how to show that it is a fundamental set of solutions on the interval. (a) and (b) confirms that it is a fundamental set of solutions, but on the interval, do i substitute in the ends of the interval? or what? i am unsure, someone please help.

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# Differential equation - 2nd order diff equation

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