If y_{1}, and y_{2} are a fundamental set of solutions of t^{2}y"-2y'+(3+t)y=0 and if W(y_{1}, y_{2})(2)=3, find the value of W(y_{1}, y_{2})(4).

Answer: 3āeā4.946

If y_{1}, and y_{2} are a fundamental set of solutions of t^{2}y"-2y'+(3+t)y=0 and if W(y_{1}, y_{2})(2)=3, find the value of W(y_{1}, y_{2})(4).

Answer: 3āeā4.946

Tutors, please sign in to answer this question.

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) = 3āeā4.946 <==Answer

Romina A.

Knowledgeable tutor in Spanish, Biology, Anatomy & Physiology and Math

Harrison, NJ

5.0
(168 ratings)

Sean B.

Premier NYC Math tutor - young, passionate and talented

New York, NY

4.8
(30 ratings)

David T.

Very Experienced, Knowledgeable, and Patient Math Tutor

Dobbs Ferry, NY

5.0
(93 ratings)

## Comments

I don't know what you wrote after W'=y

_{1}y_{2}"-y_{1}"y_{2}.Plug in y'' = (2/t^2)y' - [(3+t)/t^2]y in W',

W' = y

_{1}y_{2}"-y_{1}"y_{2}= y

_{1}{(2/t^2)y_{2}' - [(3+t)/t^2]y_{2}} - {(2/t^2)y_{1}' - [(3+t)/t^2]y_{1}}y_{2}= (2/t^2)[y

_{1}y_{2}' - y_{1}'y_{2}]= (2/t^2)W