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Find the value?

If y1, and y2 are a fundamental set of solutions of t2y"-2y'+(3+t)y=0 and if W(y1, y2)(2)=3, find the value of W(y1, y2)(4).

Answer: 3āˆšeā‰ˆ4.946

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Plug in y'' = (2/t^2)y' - [(3+t)/t^2]y in W',

W' = y1y2"-y1"y2

= y1{(2/t^2)y2' - [(3+t)/t^2]y2} - {(2/t^2)y1' - [(3+t)/t^2]y1}y2

= (2/t^2)[y1y2' - y1'y2]

= (2/t^2)W

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1 Answer

Solve for y'',

y'' = (2/t^2)y' - [(3+t)/t^2]y

W = y1y2' - y1'y2

W' = y1y2'' - y1''y2 = y1{(2/t^2)y2' - [(3+t)/t^2]y2} - {(2/t^2)y1' - [(3+t)/t^2]y1} y2

Simplify,

W' = (2/t^2) W

Separate variables,

dW/W = (2/t^2)dt

Integrate,

lnW = -2/t + c

W(t) = Ce^(-2/t)

Plug in W(2) = 3,

3 = C/e => C = 3e

W(4) = 3e e^(-2/4) = 3āˆšeā‰ˆ4.946 <==Answer