The function y ( x ) satisfies the linear equation y ″ + p (...

civilnogwu

civilnogwu

Answered

2022-07-10

The function y ( x ) satisfies the linear equation y + p ( x ) y + q ( x ) = 0.
The Wronskian W ( x ) of two independant solutions, y 1 ( x ) and y 2 ( x ) is defined as | y 1 y 2 y 1 y 2 | . Let y 1 ( x ) be given, use the Wronskian to determine a first-order inhomogeneous differential equation for y 2 ( x ). Hence show that y 2 ( x ) = y 1 ( x ) x 0 x W ( t ) y 1 ( t ) 2 d t ( )
The only first-order inhomogeneous differential equation I can get for y 2 ( x ) is simply y 1 ( x ) y 2 ( x ) y 1 ( x ) y 2 ( x ) = W ( x ). Which is trivial and simply the definition of W ( x ), so I don't think this is what it wants. I can easily get to the second result by differentiating y 2 ( x ) y 1 ( x ) and then integrating, but the wording suggests I'm supposed to get to this result from the differential equation that I can't determine.
So I would ask please that someone simply complete the above question, showing the differential equation we are to determine and how to get to the second result from this.
Edit: The question later goes on to give the differential equation x y ( 1 x 2 ) y ( 1 + x ) y = 0 call this equation ( ), we have already confirmed that y 1 ( x ) = 1 x is a solution
Hence, using ( ) with x 0 = 0 and expanding the integrand in powers of t to order t 3 , find the first three non-zero terms in the power series expansion for a solution, y 2 , of ( ) that is independent of y 1 and satisfies y 2 ( 0 ) = 0 , y 2 ( 0 ) = 1.
I'm stuck here too as I don't see how we could expand W ( t ) as it is surely a function of the unknown y 2 . Apologies for my ignorance, this is independant study, and aside from a brief look at some online notes that simply define W ( x ), I've never worked on these problems before.

Answer & Explanation

soosenhc

soosenhc

Expert

2022-07-11Added 16 answers

Your arguments sound reasonable to me, and the formulation in the exercise a bit cryptic. But you should perhaps also add that
W = | y 1 y 2 y 1 y 2 | = p W ,
from which W ( x ) = exp ( P ( x ) ) with P = p. This ensures that in ( y 2 / y 1 ) = W / y 1 2 the RHS is just an explicit function of x. Might help you on the last bit as well.

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