2MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 20We mention in passing that we have the identityW0=det°y1(t0)y2(t0)y°1(t0)y°2(t0)±°1001±=1.Linear IndependenceAbel’s Theorem.Consider the diferential equationa(t)y°°+b(t)y°+c(t)y=0.Say thaty1=y1(t) andy2=y2(t) are solutions.(1) The Wronskian oFy1andy2is the FunctionW(t)=Cexp°−²tb(τ)a(τ)dτ±For some constantC.(2)W(t)°= 0 ForalltiF and only iFW(t0)°= 0 Forsomet0.We give a prooF. Denote the FunctionW=y1y°2−y°1y2.This has the derivativeW°=³y°1y°2+y1y°°2´−³y°°1y2+y°1y°2´=y1y°°2−y°°1y2.Hence we have the expressiona(t)W°+b(t)W=a(t)µy1y°°2−y°°1y2¶+b(t)µy1y°2−y°1y2¶=a(t)µy1y°°2−y°°1y2¶+b(t)µy1y°2−y°1y2¶+c(t)µy1y2−y1y2¶=y1µa(t)y°°2+b(t)y°2+c(t)y2¶−µa(t)y°°1+b(t)y°1+c(t)y1¶y2.HenceW=W(t) satis±es a ±rst order diferential equation. Since this equation is separable, we may solveby dividing both sides byW(t):a(t)dWdt=−b(t)W1WdWdt=−b(t)a(t)ddtln|W|=−b(t)a(t)=⇒ln|W|=−²tb(τ)a(τ)dτ+ (constant).Upon exponentiating, we can expressW=W(t) in the Form
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