2nd order ODE solution bases /wronskain question

[Greger]

hello

this question from my coarse notes has been giving me some trouble so i thought i would ask for some help on here,

http://img88.imageshack.us/img88/9764/asfar.jpg

i understand that since the bases are bases of the same solutions then they are just a multiple of each other, but I'm not sure how you would show it using the wronskain.

i first tried starting by saying since both are two different bases for the solutions then
k\phi_{1}=\psi_{1}

k\phi_{2}=\psi_{2}

then doing the wronskian, but it gives k²

it seems like there is something straight forward that i am not seeing

anyone know what it might be?

 

[LCKurtz]

hello

this question from my coarse notes has been giving me some trouble so i thought i would ask for some help on here,

http://img88.imageshack.us/img88/9764/asfar.jpg

i understand that since the bases are bases of the same solutions then they are just a multiple of each other, but I'm not sure how you would show it using the wronskain.

No, they aren't just multiples of each other. For example, the equation $y''-y=0$ could have $y=A\cosh(x)+B\sinh(x)$ or $y=Ce^x+De^{-x}$.

i first tried starting by saying since both are two different bases for the solutions then
k\phi_{1}=\psi_{1}

k\phi_{2}=\psi_{2}

then doing the wronskian, but it gives k²

it seems like there is something straight forward that i am not seeing

anyone know what it might be?

What you do know is that each function in the second basis is a linear combination of the functions in the first basis. Calciulate the Wronskian for the second basis and use that fact to relate it to the Wronskian in the first basis.

 

[Greger]

Hey,

Thanks for your reply!

Sorry I had an assignment I had to work on before I started studying math again,

So I did what you said just now using

c_1\phi_{1} +c_2\phi_{2}=\psi_{1}

c_3\phi_{1} +c_4\phi_{2}=\psi_{1}

I got the required result with k = c₁c₄-c₂c₃

It seems good to me thanks!

Does it look right to you?

 

[LCKurtz]

Yes, that looks to be correct.