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, 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

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.

## 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