Do the functions #y_1(t)=√t # and #y_2(t)=1/t# form a fundamental set of solutions of the equation #2t^(2)y''+3ty'-y=0#,on interval #0<t<∞# ? Justify your answer.

1 Answer
Cesareo R.
Aug 11, 2018

See below.

Explanation:

The functions #y_1(t), y_2(t)# have similar structure so substituting into the differential equation #y(t) = t^{\alpha}# we have

# t^{alpha}(2alpha^2+alpha-1) = 0 #

and this is true for #t \ne 0# when #alpha = -1, alpha=\frac{ 1}{2}# corresponding to #y_1# and #y_2# hence #y_1# and #y_2# are a fundamental set of solutions

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