Calculus and AnalysisDetermine the values of r
2 years ago
Determine the values of r for which the given differential equation has solutions of the form y=t^(r) for t>0. t^(2)y''+4ty'+2y=0
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Best Answer
courtneybarnettticketsvectorarena Staff answered 2 years ago
Plug t^(r) into the differential equation. Recall that (d)/(dt)t^(r)=rt^(r-1) and (d^(2))/(dt^(2))t^(r)=(d)/(dt)rt^(r-1)=r(r-1)t^(r-2) t^(2)y''+4ty'+2y=0 t^(2)[r(r-1)t^(r-2)]+4t[rt^(r-1)]+2t^(r)=0 r(r-1)t^(r)+4rt^(r)+2t^(r)=0 Since t>0 we can divide both sides by t^(r) r(r-1)t^(r)+4rt^(r)+2t^(r)=0 r(r-1)+4+2=0 r^(2)-r+4r+2=0 r^(2)+3r+2=0 Solve this polynomial for r. This implies that t^(r) is a solution only if r=-1 or r=-2. r^(2)+3r+2=0 (r+1)(r+2)=0 -> r=-1, r=-2 Result: r=-1 and r=-2
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