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Differential Equation help.

Hey,

I need to show the following implication.

x is a solution to x'' + (a/t)x' + (b/t^2)x=0 iff y(u)=x(exp(u)) is a solution to y''+(a-1)y'+by=0

Where a.b are constants.

Now I have tried to use the chain rule on y(u)=x(exp(u)) then attempt to substitute and rearrange, but both the new variable u and the non constant coefficients of f(x'',x',x,t) are swaying me a bit.

Any help?

Thanks :smile:
Original post by seanylpa
...


Right, so you've got e2ux′′(eu)+aeux′(eu)+bx(eu)=0e^{2u}x''(e^u)+ae^u x'(e^u)+bx(e^u)=0

Divide by e2ue^{2u} and let t=eut=e^u etc.
Reply 2
Avatar for seanylpa
seanylpa
OP
How was that equation found?

I have managed to get to x''(exp(u))+a(x(exp(u))'+bx(exp(u))=0

Seany.
Original post by seanylpa
How was that equation found?


y(u)=x(eu)y(u)=x(e^u)

y′(u)=eux′(eu)y'(u)=e^u x'(e^u)

y′′(u)=eux′(eu)+e2ux′′(eu)y''(u)=e^{u} x'(e^u)+e^{2u} x''(e^u)
Reply 4
Avatar for seanylpa
seanylpa
OP
Ah of course :smile: Thank you :smile:

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