The wronskian of two solutions of the differential equation t2y- t(t+2)y + (t+2)y = 0 satisfies W (1) = 1 isa)t2etb)t2et-1c)t etd)t et-1Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Re-wrirte the equation as

Compare with general form y" + py' +Qy = 0, we have p(t) =

Now using W (1) = 1, we have 1 =ce i.e. c = 1/e

⇒ W(t) = t²e^t-1

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