Second Order Differential Equations ay″+by′+cy=0, without complex numbers

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shvatismop1rj · Accepted Answer

Step 1Let us show that there is a unique solution, given thatg(0)=A,g′(0)=B.Clearly Acos⁡x+Bsin⁡x is a solution with these initial conditions, so look at h(x):g(x)−Acos⁡x+Bsin⁡x; this satisfies h″+h=0, h(0)=0, h′(0)=0. We want to show that h(x)=0 for all x.Multiplying h″+h=0 by h′ we see that h″h′+h′h=0 for all x. Note that h″h′+h′h=(12h′2+12h2)′. Therefore, by the Mean Value Theorem,12h′(x)2+12h(x)2=(x−0)(12h′(θx)2+12h(θx)2=0.Now a real sum of squares can only be zero if each summand is zero. So we have, as required, h(x)=0 for all x.[For those of an applied bent, we are just doing the classical thing of conserving the total energy.]

Jazlyn Mitchell · Accepted Answer

Step 1You can make the educated guess that the solution will be of the formf(x)=ekxg(x).Then we geta[ekxg(x))″+b(ekxg(x))′+cekxg(x)=a[ekxg″(x)+2kekxg′(x)+k2ekxg(x)]+b[ekxg′(x)+kekxg(x)]+cekxg(x)=aekxg″(x)+[2ak+b]ekxg′(x)+[ak2+bk+c]ekxg(x)=0Since ekx is never 0, this reduces toag″+(2ak+b)g′+(ak2+bk+c)g=0.We are free to choose k as we wish, because if ekxg(x) solves the ODE, then so does ek~xe(k−k~)xg(x), and we simply get g(x)~=e(k−k~)g(x) as the solution for g instead. So we choose k as simple as possible: such that 2ak+b=0, that is, k=−b2a. Then the equation reads (after multiplying by 4a)4a2g″−(b2−4ac)g=0.The solution to this, given that b2−4ac&amp;lt;0, is known: it&#039;s of the form g(x)=Asin⁡(ωx)+Bcos⁡(ωx), where ω=c−b24a.So in total we get f(x)=ekx[Asin⁡(ωx)+Bcos⁡(ωx)] with k and ω as specified.

Second Order Differential Equations \(\displaystyle{a}{y}{''}+{b}{y}'+{c}{y}={0}\), without complex

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