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
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
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
Comment