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Differential EquationsHardThe Wronskian & Variation of Parameters

The Wronskian & Variation of Parameters · Hard problem

One solution of (1−t2)y′′−2ty′+2y=0(1-t^2)y''-2ty'+2y=0 on −1<t<1-1<t<1 is y1=ty_1=t. Find the solution with y(0)=1y(0)=1 and y′(0)=0y'(0)=0, and give y ⁣(12)y\!\left(\dfrac12\right).